Answer:
a) Null hypothesis:[tex]\mu \leq 30.2[/tex]
Alternative hypothesis:[tex]\mu > 30.2[/tex]
b) [tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]
And the distribution for the random sample is given by:
[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]
c) [tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]
[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]
d) If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.
e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=32.12[/tex] represent the sample mean
[tex]s=4.83[/tex] represent the sample standard deviation
[tex]n=50[/tex] sample size
[tex]\mu_o =30.2[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
a. Define the parameter and state the hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 30.2[/tex]
Alternative hypothesis:[tex]\mu > 30.2[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
b. Define the sampling distribution (mean and standard deviation).
Let X the random variable who represent the variable of interest. And we know that the distribution for X is:
[tex]X \sim N(\mu=32.12, \sigma=4.83)[/tex]
And the distribution for the random sample is given by:
[tex]\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)[/tex]
c. Perform the test and calculate P-value
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=50-1=49[/tex]
Since is a one right tailed test the p value would be:
[tex]p_v =P(t_{(49)}>2.81)=0.0035[/tex]
d. State your conclusion.
If we compare the p value and the significance level assumed [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.
e. Explain what the p-value means in this context.
The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.
Final answer:
Explanation of hypothesis testing for a fuel economy standard compliance claim.
Explanation:
Hypotheses:
Parameter: Average fuel economyNull Hypothesis (H0): μ = 35.5Alternative Hypothesis (Ha): μ < 35.5Sampling Distribution:
Mean: 34.6 mpgStandard Deviation: 10.3 mpgPerforming the Test:
Calculate the Z-score and P-value from the sample data using the formula. Compare the P-value to the significance level.
Conclusion: Since **p-value = 0.2578** is greater than the significance level of 0.05, we fail to reject the null hypothesis. There is not enough evidence to claim that the fleet meets the fuel economy standard.
More people are using social media to network, rather than phone calls or e-mails (US News & World Report, October 20, 2010). From an employment perspective, jobseekers are no longer calling up friends for help with job placement, as they can now get help online. In a recent survey of 150 jobseekers, 67 said they used LinkedIn to search for jobs. A similar survey of 140 jobseekers, conducted three years ago, had found that 58 jobseekers had used LinkedIn for their job search. Use Table 1.
Let p1 represent the population proportion of recent jobseekers and p2 the population proportion of job seekers three years ago. Let recent survey and earlier survey represent population 1 and population 2, respectively.a. Set up the hypotheses to test whether there is sufficient evidence to suggest that more people are now using LinkedIn to search for jobs as compared to three years ago.
A. H0: p1 − p2 ≥ 0; HA: p1 − p2 < 0
B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
C. H0: p1 − p2 = 0; HA: p1 − p2 ≠ 0
b. Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)c. Calculate the critical value at the 5% level of significance. (Round your answer to 3 decimal places.)d. Interpret the results.
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
B. Do not reject H0; there is an increase in the proportion of people using LinkedIn
C. Reject H0; there is no increase in the proportion of people using LinkedIn
D. Reject H0; there is an increase in the proportion of people using LinkedIn
Answer:
B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
[tex]z=\frac{0.447-0.414}{\sqrt{0.431(1-0.431)(\frac{1}{150}+\frac{1}{140})}}=0.57[/tex]
[tex]z_{crit}=1.64[/tex]
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
Step-by-step explanation:
1) Data given and notation
[tex]X_{1}=67[/tex] represent the number of recent jobseekers
[tex]X_{2}=58[/tex] represent the number of job seekers three years ago.
[tex]n_{1}=150[/tex] sample of recent jobseekers selected
[tex]n_{2}=140[/tex] sample of job seekers three years ago selected
[tex]p_{1}=\frac{67}{150}=0.4468[/tex] represent the proportion of recent jobseekers
[tex]p_{2}=\frac{58}{140}=0.4143[/tex] represent the proportion of job seekers three years ago
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.05[/tex] significance level given
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if "More people are using social media to network, rather than phone calls or e-mails", the system of hypothesis would be:
Null hypothesis:[tex]p_{1} - p_{2} \leq 0[/tex]
Alternative hypothesis:[tex]p_{1} - p_{2} > 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{67+58}{150+140}=0.4310[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.4468-0.4143}{\sqrt{0.4310(1-0.4310)(\frac{1}{150}+\frac{1}{140})}}=0.5671[/tex]
In order to find the critical value since we have a right tailed test the we need to find a value on the z distribution that accumulates 0.05 of the area on the right tail, and this value is[tex]z_{crit}=1.64[/tex].
4) Statistical decision
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>0.5671)= 0.285[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
So the correct conclusion would be:
A. Do not reject H0; there is no increase in the proportion of people using LinkedIn
Students in a discussion of gun control in a sociology class at Foothill Community College argue that Republicans are more likely to oppose gun control than Independents. They use data from an article titled "Gun Control Splits America," published March 23, 2010 in pewresarch.org by the Pew Research Center for the People and the Press. In this study 62% of Republicans and 57% of Independents say that states should not be able to pass laws banning handguns.
For a claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents, the null and alternative hypotheses are
H0: p1-p2 = 0 (p1 = p2)
Ha: p1-p2 > 0 (p1 > p2)
The p -value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion?
A. Reject H0 , and support Ha.
B. Support H0 , and reject Ha.
C. Fail to Reject H0.
D. do not support Ha .
Answer:
C. Fail to Reject H0.
Step-by-step explanation:
If the P-value is 0.06, that means that the result enters in the acceptance region. It is a value that is expected to happen if both proportions are equal (null hypothesis) at this significance level.
The conclusion when the P-value is bigger than the significance level is that the effect is not significant and it failed to reject the null hypothesis.
Final answer:
The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are H0: p1-p2 = 0 (p1 = p2) and Ha: p1-p2 > 0 (p1 > p2).
The p-value of 0.06 is larger than the significance level of 0.05, so the appropriate conclusion is C. Fail to Reject H0.
Explanation:
The null and alternative hypotheses for the claim that a larger proportion of Republicans oppose state laws banning handguns when compared to Independents are:
H0: p1-p2 = 0 (p1 = p2)
Ha: p1-p2 > 0 (p1 > p2)
The p-value of 0.06 is larger than the significance level of 0.05. Therefore, we fail to reject the null hypothesis. Hence, an appropriate conclusion is:
C. Fail to Reject H0.
The net weights (in grams) of a sample of bottles filled by a machine manufactured by Edne, and the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are; Edne 8 7 6 9 7 5 Orno 10 7 11 9 12 14 9 8 Testing the claim at the 0.05 level that the mean weight of the bottles filled by the Orno machine is greater than the mean weight of the bottles filled by the Edne machine, what is the critical value for this test?
Answer:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 \leq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 > \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 \leq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 > 0[/tex]
Our notation on this case :
[tex]n_1 =8[/tex] represent the sample size for group 1 (Orno)
[tex]n_2 =6[/tex] represent the sample size for group 2 (Edne)
We can calculate the sampel means and deviations with the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X_1 =10[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =7[/tex] represent the sample mean for the group 2
[tex]s_1=2.268[/tex] represent the sample standard deviation for group 1
[tex]s_2=1.414[/tex] represent the sample standard deviation for group 2
If we see the alternative hypothesis we see that we are conducting a right tailed test.
On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution [tex]\alpha/2 = 0.05[/tex] of the area.
The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=8+6 -2=12[/tex] degrees of freedom.
We can use the following excel code in order to find the critical value:
"=T.INV(1-0.05,12)"
And the rejection zone is: (1.78,infinity)
The statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]
The degrees of freedom are given by:
[tex]df=8+6-2=12[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).
A contractor claims that their soundproofing will remove 83% of the sound intensity inside the room. If 83% of the sound intensity inside the room is removed, the new sound level will sound what % less loud to people in the room? Round your answer to the nearest 1%.
Answer:
17%
Step-by-step explanation:
If 83% is removed than
total is always 100%
that's why
so the new sound level will sound
100%-83%=17% remains.
hence 17 % is correct answer .
All fifth-grade students are given a test on academic achievement in New York State. Suppose the mean score is 70 for the entire state. A random sample of fifth-grade students is selected from Long Island. Below are the scores in this sample from a normal population. 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74 a. Construct a 95% confidence interval for the population mean score on Long Island. b. Construct a 90% confidence interval for the population standard deviation of the scores on Long Island. c. A teacher at a Long Island high school claims that the mean score on Long Island is higher than the mean for New York State. Conduct a test to see if this claim is reasonable using α = 0.01. d. Find the p-value of the test.
Answer:
a) The 90% confidence interval would be given by (73.56;79.84)
b) The 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].
c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance. So the claim of the teacher makes sense.
d) Since is a one-side upper test the p value would given by:
[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]
Step-by-step explanation:
Data given: 82 94 66 87 68 85 68 84 70 83 65 70 83 71 82 72 73 81 76 74
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
We can calculate the sample mean and deviation with the following formulas:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And we got the following results:
[tex]\bar X=76.7[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=8.112[/tex] represent the population standard deviation
n=20 represent the sample size
90% confidence interval
Part a
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,19)".And we see that [tex]t_{\alpha/2}=1.73[/tex]
Now we have everything in order to replace into formula (1):
[tex]76.7-1.73\frac{8.112}{\sqrt{20}}=73.56[/tex]
[tex]76.7+1.73\frac{8.112}{\sqrt{20}}=79.84[/tex]
So on this case the 90% confidence interval would be given by (73.56;79.84)
Part b
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=30.143[/tex]
[tex]\chi^2_{1- \alpha/2}=10.117[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(8.112^2)}{30.143} \leq \sigma^2 \leq \frac{(19)(8.112^2)}{10.117}[/tex]
[tex] 41.472 \leq \sigma^2 \leq 123.574[/tex]
So the 90% confidence interval for the deviation would be [tex] 6.44 \leq \sigma \leq 11.116[/tex].
Part c
Null hypothesis:[tex]\mu \leq 70[/tex]
Alternative hypothesis:[tex]\mu > 70[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{76.7-70}{\frac{8.112}{\sqrt{20}}}=3.69[/tex]
Part d
P-value
Since is a one-side upper test the p value would given by:
[tex]p_v =P(t_{19}>3.69)=0.00078[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 70 at 1% of significance.
Answer:
(a) 95% confidence interval for the population mean score on Long Island is (73, 80.40)
(b) 90% confidence interval for the population standard deviation of the scores on Long Island is (4.85, 10.97)
(c) The teacher's claim that the mean score on Long Island is higher than the mean for New York State is reasonable
(d) p-value is 0.01
Step-by-step explanation:
From the data values from Long Island,
Mean is 76.7 and standard deviation is 7.91
Confidence Interval (CI) = mean + or - (t × sd)/√n
(a) mean = 76.7, sd = 7.91, n = 20, degree of freedom = n-1 = 20-1 = 19, t-value corresponding to 19 degrees of freedom and 95% confidence level is 2.093
Lower bound = 76.7 - (2.093×7.91)/√20 = 76.7 - 3.70 = 73
Upper bound = 76.7 + (2.093×7.91)/√20 = 76.7 + 3.70 = 80.40
95% CI is (73, 80.40)
(b) CI = sd + or - (t×sd)/√n
sd = 7.91, t-value corresponding to 19 degrees of freedom and 90% confidence level is 1.729
Lower bound = 7.91 - (1.729×7.91)/√20 = 7.91 - 3.06 = 4.85
Upper bound = 7.91 + (1.729×7.91)/√20 = 7.91 + 3.06 = 10.97
90% CI is (4.85, 10.97)
(c) Null hypothesis: The mean score on Long Island is 70
Alternate hypothesis: The mean score on Long Island is greater than 70
Z = (sample mean - population mean)/(sd/√n) = (70 - 76.7)/(7.91/√20) = -6.7/1.77 = -3.79
Using 0.01 significance level, the critical value is 2.326
Since -3.79 is less than 2.326, reject the null hypothesis. The teacher's claim is reasonable
(d) p-value = 1 - cumulative area of test statistic = 1 - 0.9900 = 0.01
Evaluate the expression for the given values of the variables.
Evaluate 9p − 8q for p = 4 and q = −8.
The expression is equal to .
The expression 9p − 8q for p = 4 and q = −8 is equal to 100
Solution:Given that we have to evaluate expression for the given values of the variables
Expression is:
⇒ 9p − 8q
Given that p = 4 and q = -8
Let us substitute the given values of p = 4 and q = -8 in given expression and evaluate it
⇒ 9p − 8q = 9(4) - 8(-8)
⇒ 9p - 8q = 9(4) + 8(8)
Upon multiplying the terms we get,
⇒ 9p - 8q = 36 + 64 = 100
Thus the expression is equal to 100
A 20 year par value bond with semi-annual coupons at a nominal annual rate of 8% convertible semi-annually is purchased at a price of 1783.27. The bond can be called at par value X on any coupon date starting at the end of year 12 after the coupon is paid. The price guarantees a nominal annual rate of interest convertible semi-annually of at least 6%. Calculate X.
Answer:
3.216%
Step-by-step explanation:
This bond sells at a higher price or value, which means that its coupon is bogus of market interest rate. Therefore, the minimum yield rate that accounts for the possibility of the bond being called is calculated at the earliest possible call date. Let say exactly 15 years from the date of purchase, because that would be the most disadvantageous date for the bondholder for the call to occur.
The minimum semiannual yield:
j= i²/2
i² = 2j
which therefore satisfies the expression below for the worst possible case scenario yield:
1722.25 = 0.04*1100*[tex]a]_30[/tex]+[tex]\frac{1100}{(1+j)^30}[/tex]
Also, with the use of a financial calculator (making sure that the calculator is not in BGN mode)
1722.25 PV, -44 PMT, -1100 FV, 30 N, CPT 1/Y.
j can be found to be 1.608245%. The corresponding nominal annual rate compounded semiannually is (X) = i² = 2j =3.216%
Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximations to the integral using n and 2n subintervals. b. Find the Simpson's rule approximation to the integral using 2n subintervals. c. Compute the absolute errors in the trapezoid rule and Simpson's rule with 2n subintervals.
Answer:
a.
With n = 25, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549[/tex]
With n = 50, [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594[/tex]
b. [tex]\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943[/tex]
c.
The absolute error in the trapezoid rule is 0.08047
The absolute error in the Simpson's rule is 0.00008
Step-by-step explanation:
a. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 25 with the trapezoid rule you must:
The trapezoidal rule states that
[tex]\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We have that a = 0, b = 1, n = 25.
Therefore,
[tex]\Delta{x}=\frac{1-0}{25}=\frac{1}{25}[/tex]
We need to divide the interval [0,1] into n = 25 sub-intervals of length [tex]\Delta{x}=\frac{1}{25}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579[/tex]
...
[tex]2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701[/tex]
[tex]f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the trapezoid rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549[/tex]
To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using n = 50 with the trapezoid rule you must:We have that a = 0, b = 1, n = 50.
Therefore,
[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]
We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
...
[tex]2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705[/tex]
[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the trapezoid rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594[/tex]
b. To approximate the integral [tex]\int_{0}^{1}e^{6 x}\ dx[/tex] using 2n with the Simpson's rule you must:
The Simpson's rule states that
[tex]\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We have that a = 0, b = 1, n = 50
Therefore,
[tex]\Delta{x}=\frac{1-0}{50}=\frac{1}{50}[/tex]
We need to divide the interval [0,1] into n = 50 sub-intervals of length [tex]\Delta{x}=\frac{1}{50}[/tex], with the following endpoints:
[tex]a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b[/tex]
Now, we just evaluate the function at these endpoints:
[tex]f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1[/tex]
[tex]4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175[/tex]
[tex]2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281[/tex]
...
[tex]4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541[/tex]
[tex]f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735[/tex]
Applying the Simpson's rule formula we get
[tex]\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943[/tex]
c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by [tex]|A-B|[/tex]
The absolute error in the trapezoid rule is
The calculated value is
[tex]\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225[/tex]
and our estimate is 67.1519320308594
Thus, the absolute error is given by
[tex]|67.0714655821225-67.1519320308594|=0.08047[/tex]
The absolute error in the Simpson's rule is
[tex]|67.0714655821225-67.0715427161943|=0.00008[/tex]
ureg placed 32 chairs in the auditorium. There are 8 chairs in each row. Which
equation could be used to represent this situation
A. 32 x 8
00
--
B. 32 + 8
00
8 = 32
D.
00
x 8 = 32
The equation to represent this situation is 8x = 32
Solution:
Given that ureg placed 32 chairs in the auditorium
There are 8 chairs in each row
To find: equation used to represent this situation
From given information,
1 row = 8 chairs
So let us find an expression to determine the number of rows to place 32 chairs
Let "x" be the number of rows required to place 32 chairs
Since 1 row contains 8 chairs, expression to determine the number of rows to place 32 chairs is given as:
[tex]8 \times \text{ number of row } = 32[/tex]
8x = 32
Thus the equation to represent this situation is 8x = 32
Ray Flagg took out a 60-month fixed installment loan of $12,000 to open a new pet store. He paid no money down and began making monthly payments of $232. Ray's business does better than expected and instead of making his 30th payment, Ray wishes to repay his loan in full.
Answer:
Ray Flagg will pay $5,272 at the time of his 30th installment.
Step-by-step explanation:
Ray took $12,000 load for 60 months. As he paid no amount as down payment so his monthly payment will be $200:
[tex]=12000/60\\=200[/tex]
Instead of $200 per month, he used to pay $232 per month. So, before his 30th installment, he paid 29 installments each of $232 which is $6,728:
[tex]=232*29\\=6728[/tex]
As the business does better, he wishes to payback remaining amount at once so he will pay $5,272 as:
[tex]12000-6728\\=5272[/tex]
To calculate the remaining balance of a fixed installment loan after a certain number of payments, use the formula provided in the detailed answer.
Explanation:Mathematics: High SchoolRay Flagg took out a 60-month fixed installment loan of $12,000 to open a new pet store. He paid no money down and began making monthly payments of $232. Ray's business does better than expected and instead of making his 30th payment, Ray wishes to repay his loan in full.
To calculate the remaining balance after 29 months, we can use the formula for the remaining balance of a fixed installment loan:
Remaining Balance = Balance × (1 + Monthly Interest Rate)Number of Payments Made - (Monthly Payment × ((1 + Monthly Interest Rate)Number of Payments Made - 1) / Monthly Interest Rate)
Using the given values, the monthly interest rate can be calculated by dividing the annual interest rate by 12 and converting it to a decimal.
Finally, substitute the values into the formula to find the remaining balance after 29 months.
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The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (Round your answer to two decimal places.)
Answer:
If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 15
Standard Deviation, σ = 1
Sample size = 4
Total lifetime of 4 batteries = 40 hours
We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5[/tex]
We have to find the value of x such that the probability is 0.05
P(X > x) = 0.05
[tex]P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95 [/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83[/tex]
Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.
A telemetry voltage V, transmitted from a position sensor on a ship's rudder, is a random variable with PDF:
fV(v)={1/32 0 −16<_v<_16, otherwise.
A receiver in the ship's control room receive R = V+X. The random variable X is a Gaussian (4,4) noise voltage that is independent of V. The receiver uses R to calculate a linear estimate of the telemetry voltage: V = aR+b.
(a) Find the expected value of the received voltage. E [R] = _______
(b) Find the variance of the received voltage. Var [R] = _________
(c) Find the covariance of the transmitted and received voltage. Cov [V, R] = _________
(d) Find the optimal linear estimate. VL (R) = __________
(e) Compute the minimum mean square error of the estimate. e* = __________
The expected value of the received voltage is 4. The variance of the received voltage is 64/3. The covariance of the transmitted and received voltage is 0.
(a) To find the expected value of the received voltage, we need to use the linearity property of the expectation and the fact that V and X are independent. The expected value of R is given by:
E[R] = E[V+X] = E[V] + E[X]
Since V and X are independent, we have E[X] = 4 and E[V] = 0 (by symmetry of the uniform distribution). Therefore, E[R] = 0 + 4 = 4.
(b) To find the variance of the received voltage, we can use the properties of variance. Variance is additive for independent random variables, so:
Var[R] = Var[V+X] = Var[V] + Var[X]
Since V and X are independent, we have Var[X] = 4^2 = 16 and Var[V] = (16^2)/12 = 64/12 = 16/3. Therefore, Var[R] = 16/3 + 16 = 64/3.
(c) The covariance of the transmitted and received voltage is given by:
Cov[V, R] = E[(V - E[V])(R - E[R])]
Since E[V] = 0 and E[R] = 4, this simplifies to:
Cov[V, R] = E[VR] - E[V]E[R]
Since V and R are independent, we have Cov[V, R] = E[V]E[R] - E[V]E[R] = 0.
(d) The optimal linear estimate VL(R) is given by:
VL(R) = E[V] + Cov[V, R]/Var[R] * (R - E[R])
Since Cov[V, R] = 0, the optimal linear estimate becomes:
VL(R) = E[V] + 0/Var[R] * (R - E[R]) = E[V] = 0.
(e) The minimum mean square error of the estimate is given by:
e* = Var[V - VL(R)]
Since VL(R) = E[V] = 0, this simplifies to:
e* = Var[V] = 16/3.
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Determine the horizontal change of a line with an x intercept at (3,0) and a y intercept at (0,2)
Answer:the horizontal change of the line is 3
Step-by-step explanation:
The horizontal change is the change in the value of x on the horizontal axis. It is expressed as
x2 - x1
Where
x2 represents the final value of x
x1 represents the initial value of x
An x intercept at (3,0) means that the line cut across the x axis at the point when x = 3 and y = 0
A y intercept at (0,2) means that the line cut across the y axis at the point when 0 = 3 and y = 2
Change in the horizontal axis would be x2 - x1 = 3 - 0 = 3
Errors in measuring the time of arrival of a wave front from an acoustic source sometimes have an approximate beta distribution. Suppose that these errors, measured in microseconds, have a approximately a beta distribution with α = 1 and β = 2
What is the probability that the measurement error in a randomly selected instance us less than 0.6 µs?
Give the mean and standard deviation of the measurement errors.
Answer:
a) P=0.84
b) Mean=0.33
Standard deviation=0.356
Step-by-step explanation:
The probabilty that the measurement error in a randomly selected instance us less than 0.6 µs is P=0.84.
The mean of a Beta(α = 1, β = 2) is
[tex]\mu=\frac{\alpha}{\alpha+\beta}=\frac{1}{1+2}=0.33[/tex]
The standard deviation of a Beta(α = 1, β = 2) is
[tex]\sigma=\sqrt{\frac{\alpha\beta}{(\alpha+\beta)^2*(\alpha+\beta+1)}}\\\\\\\sigma= \sqrt{\frac{1*2}{(1+2)^2*(1+2+1)}}=\sqrt{\frac{2}{(2)^2*(4)}}=\sqrt{\frac{2}{16} } =\sqrt{0.125}= 0.356[/tex]
Given a Beta Distribution with α = 1 and β = 2, probability calculations typically rely on integration of the probability density function, often executed with statistical software. The mean and standard deviation can be mathematically derived with the given α and β, giving us mean= 0.33 and standard deviation= 0.236.
Explanation:The question falls under the domain of probability distribution, specifically the Beta Distribution. The Beta distribution is a family of continuous probability distributions defined on the interval [0, 1] parameterized by two positive shape parameters, denoted by α and β. For a Beta distribution, the probability density function is given by f(x; α, β). When α = 1 and β = 2, the beta distribution becomes a decreasing linear function.
To find the probability that the measurement error is less than 0.6, we need to integrate the probability density function from 0 to 0.6. However, it's important to note that directly performing these integrations and calculations is a task typically performed by statistical software.
Concerning mean and standard deviation of the Beta Distribution, the mean (µ) is given by α / (α + β), and the variance (σ²) is given by (αβ) / [(α+β)²*(α+β+1)]. Here, given α=1 and β=2, we find µ=1/(1+2) = 0.33 and σ² = (1*2)/(3²*4)=0.0556, and taking square root of variance we get standard deviation σ = 0.236.
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Many couples believe that it is getting too expensive to host an "average" wedding in the United States. According to the website www.costofwedding, the average cost of a wedding in the U.S. in 2009 was $24,066. Recently, in a random sample of 40 weddings in the U.S. it was found that the average cost of a wedding was $23,224, with a standard deviation of $2,903. On the basis of this, a 95% confidence interval for the mean cost of weddings in the U.S. is $22,296 to $24,152.
Answer:
Step-by-step explanation:
Given that many couples believe that it is getting too expensive to host an "average" wedding in the United States.
Population mean =24066
Sample mean = 23224
Sample size = 40
Sample std dev = 2903
Since sample std dev is known, we use t critical value.
df =39
Sample mean follows a normal distribution with mean = 23224, and std dev = [tex]\frac{s}{\sqrt{n} } \\=\frac{2903}{\sqrt{40} } \\=459.005[/tex]
t critical value = 2.023
Margin of error = 2.023*459.005
Confidence interval[tex](22295.43, 24152.57)[/tex]
Final answer:
The question involves calculating a 95% confidence interval for the average cost of weddings in the U.S., resulting in a range of $22,296 to $24,152, based on recent sample data.
Explanation:
The question pertains to the formation of a confidence interval for the average cost of weddings in the U.S. based on a sample. The provided data asserts that the average cost of a wedding in 2009 was $24,066, while a more recent sample reveals an average of $23,224 with a standard deviation of $2,903. Calculating a 95% confidence interval results in a range of $22,296 to $24,152. This implies that we can be 95% confident that the true average cost of weddings in the whole population falls within this interval.
A psychiatrist studying the affects of healthy eating habits on mood improvement. He identifies 25 people who eat healthy food and 25 who do not. Each other 50 people is given a questionnaire design to determine their mood. None of the 50 people participated in the study knew they were part of the study. Which statement is true? A this is a random mise comparative experiment be this is a double blind study see this is an observational study D this is matched pairs design
Answer:
The statement which is true is:
B. This is a double blind study
Step-by-step explanation:
Double Blind Study is such a study method in which participants don't know about the information that can effect the participants in order to eliminate the bias. This is the case with out situation in which participants didn't know that they were the part of a study. Randomized Comparative Experiment is such an experiment in which participants are randomly selected for different treatments as well as the comparison of effects of different treatments is done. So, the option A is not valid in our situation.Observational Study is such a study in which we only observe the the data collected or participants. We observe the cause-effect relationship in this study. Matched Pair Design is such an experiment in which subjects are grouped in pairs. It is a special type of randomized block design.Answer:
C. Observational Study
Step-by-step explanation:
There arent any treatments being manipulated so it can't be an experiment. The researcher is aware of the experiment being performed so its not double-blind. The only option it could be is C, because the researcher is observing the people eating and isn't manipulating their eating habits in any way.
c. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
Answer:
The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.
Step-by-step explanation:
Q. This is the question:The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
Answer:This is the question of normal distribution:
First w calculate the value of Z corresponding to X = 60,000 miles
We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles
Now, for Z, we know that:
Z = (x-μ)/σ
Z = (60,000 - 50,000)/5,000
Z = 2
Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.
P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)
P(X > 60,000) = 1 - 0.9772
P(X > 60,000) = 0.0228
For each of the given situations, write out the alternative hypothesis, being sure to state whether it is one-sided or two-sided.a) A consumer magazine discovered that 30% of a certain computer model had warranty problems over the first three months. From a random sample, the manufacturer wants to know if a new model has improved that rate.Complete the alternative hypothesis and determine whether the alternative hypothesis is one-sided or two-sided.
Answer:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
Step-by-step explanation:
1) Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
2) Solution to the problem
On this case we want to test is [tex]p>0.3[/tex] since we want to check if the new model has improved the warranty rate, we can express it like this:
[tex]p-0.3<0[/tex] since are equivalent expressions.
And the alternative hypothesis should be the complement:
Null hypothesis: [tex]p\leq 0.3[/tex] or [tex]p=0.3[/tex]
So the correct system of hypothesis for this case would be:
Null hypothesis: [tex]p\leq 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
The alternative way should be:
Null hypothesis: [tex]p = 0.3[/tex]
Alternative hypothesis: [tex]p > 0.3[/tex]
A farmer wishes to test the effects of a new fertilizer on her tomato yield. She has four equal-sized plots of land-- one with sandy soil, one with rocky soil, one with clay-rich soil, and one with average soil. She divides each of the four plots into three equal-sized portions and randomly labels them A, B, and C. The four A portions of land are treated with her old fertilizer. The four B portions are treated with the new fertilizer, and the four C's are treated with no fertilizer. At harvest time, the tomato yield is recorded for each section of land. What type of experimental design is this? completely randomized design double-blind design matched-pairs design randomized block design
Answer:
Consider the following explanation
Step-by-step explanation:
Completely Randomized Design :-
A completely randomized design is probably the simplest experimental design, in terms of data analysis and convenience. With this design, subjects are randomly assigned to treatments.
This completely randomized design relies on randomization to control for the effects of extraneous variables. The experimenter assumes that, on average, extraneous factors will affect treatment conditions equally; so any significant differences between conditions can fairly be attributed to the independent variable.
Double Blind Design :-
In an experiment, if subjects in the control group know that they are receiving a placebo, the placebo effect will be reduced or eliminated; and the placebo will not serve its intended control purpose.
Blinding is the practice of not telling subjects whether they are receiving a placebo. In this way, subjects in the control and treatment groups experience the placebo effect equally. Often, knowledge of which groups receive placebos is also kept from analysts who evaluate the experiment. This practice is called double blinding. It prevents the analysts from "spilling the beans" to subjects through subtle cues; and it assures that their evaluation is not tainted by awareness of actual treatment conditions.
Matched Pairs Design :-
A matched pairs design is a special case of a randomized block design. It can be used when the experiment has only two treatment conditions; and subjects can be grouped into pairs, based on some blocking variable. Then, within each pair, subjects are randomly assigned to different treatments.
Randomized Block Design :-
With a randomized block design, the experimenter divides subjects into subgroups called blocks, such that the variability within blocks is less than the variability between blocks. Then, subjects within each block are randomly assigned to treatment conditions. Compared to a completely randomized design, this design reduces variability within treatment conditions and potential confounding, producing a better estimate of treatment effects.
The experimental design that is given in the problem, is an example of a Randomized Block Design. Here, the different type of soils can be considered different blocks. In these blocks, the variability among within the blocks is minimum and between the blocks, it is maximum. Also, the treatments ( fertilizers ) are assigned randomly to the plts in each block (different soil type ).
Daniele has 33 quarters and dimes in her piggy bank. The piggy bank contains a total of $4.95. Write and solve a system of equations to find the number of dimes x and the number of quarters y. Solve the linear system by substitution.
Answer:the number of dimes is 22
the number of quarters is 11
Step-by-step explanation:
A dime is worth 10 cents. Converting to dollars, it becomes 10/100 = $0.1
A quarter is worth 25 cents. Converting to dollars, it becomes 25/100 = $0.25
Let x represent the number of dimes.
Let y represent the number of quarters.
Daniele has 33 quarters and dimes in her piggy bank. This means that
x + y = 33
The piggy bank contains a total of $4.95. This means that
0.1x + 0.25y = 4.95 - - - - - - - - -1
Substituting x = 33 - y into equation 1, it becomes
0.1(33 - y) + 0.25y = 4.95
3.3 - 0.1y + 0.25y = 4.95
- 0.1y + 0.25y = 4.95 - 3.3
0.15y = 1.65
y = 1.65/0.15 = 11
Substituting y = 11 into x = 33 - y, it becomes
x= 33 - 11 = 22
Final answer:
To find the number of dimes and quarters in Daniele's piggy bank, a system of equations can be solved by substitution method.
Explanation:
System of Equations:
Let x be the number of dimes and y be the number of quarters.
We have the equations: 0.10x + 0.25y = 4.95 and x + y = 33.
Solving by substitution, we first solve x + y = 33 for x to get x = 33 - y.
Substitute x = 33 - y into 0.10x + 0.25y = 4.95 and solve to find the values of x and y.
The solution is x = 15, y = 18, so there are 15 dimes and 18 quarters.
The incorrect work of a student to solve an equation 2(y + 8) = 4y is shown below:
Step 1: 2(y + 8) = 4y
Step 2: 2y + 10 = 4y
Step 3: 2y = 10
Step 4: y = 5
Which of the following explains how to correct Step 2 and shows the correct value of y? (5 points)
Group of answer choices
Answer:
Step 2 involved distributive property and the value of y is equal to 8.
Step-by-step explanation:
In step 2, there has been an error in applying the Distributive Property correctly.
Distributive Property- a(b+c)
= a x b + a x c
Step 2: [tex]2y+16=4y[/tex]
Step 3:[tex]2y=16[/tex]
Step 4: [tex]y=8[/tex]
y=8
Answer:
2 should be distributed as 2y + 16; y + 8
a chemist needs 120 milliliters of a 72% solution but has only 51% and 87% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
Answer:
The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.
Step-by-step explanation:
If x is the volume of 51% solution, and y is the volume of 87% solution, then:
x + y = 120
0.51x + 0.87y = 0.72(120)
Solve the system of equations.
0.51x + 0.87(120 − x) = 0.72(120)
0.51x + 104.4 − 0.87x = 86.4
0.36x = 18
x = 50
y = 70
The chemist needs 50 mL of 51% solution and 70 mL of 87% solution.
Antonette gets $70\%$ on a 10-problem test, $80\%$ on a 20-problem test and $90\%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is her overall score, rounded to the nearest percent?
Answer:
Percentage score will be 83.33 %
Step-by-step explanation:
We have given Antonette gets 70 % on 10 problem test
Let consider here here total problem = total marks
So marks get 10 10 problem test = 10×0.7 = 7
Marks get in 20 problem test = 20×0.8 = 16
And marks get in 30 problem test = 30×0.9 = 27
Now total marks get get by Antonette = 7 +16 + 27 = 50
And total marks = 60
So percentage score of Antonette [tex]=\frac{50}{60}\times 100=83.33[/tex] %
A consumer activist group wants to determine the mean lifetime of the Amazon Kindle DX. The consumer activist groups randomly selects 25 Kindles and finds that the average lifespan was 38 months with standard deviation 12 months. Find a 95% confidence interval for the population mean lifetime of the Amazon Kindle DX.
Answer:
Confidence interval for the population mean lifetime of the Amazon Kindle DX is (33.30 months to 42.70 months)
Step-by-step explanation:
Given;
Mean lifespan x = 38 months
Standard deviation r = 12 months
Number of kindle selected n = 25
Confidence range = 95%
Z*(95%) = 1.96
Confidence interval = x+/-Z*(r/√n)
= 38 +/- 1.96(12/√25)
= 38 +/- 4.70
Confidence interval = (33.30 months to 42.70 months)
The average maximum monthly temperature in a city is 29.9 degrees Celsius. The standard deviation in maximum monthly temperature is 2.31 degrees. Assume that maximum monthly temperatures are normally distributed. Use this Rule of Thumb to complete the sentence. Round your answers to one decimal place. (Enter your answers from smallest to largest.) 95% of the time the maximum monthly temperature is between and degrees Celsius.
Answer: 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.
Step-by-step explanation:
The Range Rule of Thumb tells that the range is approximately four times the standard deviation.
95% of the data lies within 2 standard deviations from the mean .
Maximum usual value = Mean +2 (Standard deviation )
Minimum usual value = Mean - 2 (Standard deviation)
Given : Mean = 29.9 degrees Celsius
Standard deviation = 2.31 degrees Celsius
Then, according to the Range Rule of Thumb , we have
Maximum usual value = 29.9 +2 (2.31) = 34.52 degrees Celsius
Minimum usual value = 29.9 - 2 (2.31) = 25.28 degrees Celsius
i.e. 95% of the time the maximum monthly temperature is between 25.28 degrees Celsius and 34.52 degrees Celsius.
Using the empirical rule for normal distribution, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius, given the mean is 29.9°C and the standard deviation is 2.31°C.
Explanation:The question asks for the range of temperatures within which 95% of the maximum monthly temperatures fall, given that the average maximum monthly temperature is 29.9 degrees Celsius with a standard deviation of 2.31 degrees, assuming a normal distribution. To find this, we can apply the empirical rule that states approximately 95% of the data in a normal distribution falls within two standard deviations of the mean. Therefore:
Calculate the lower boundary by subtracting two standard deviations from the mean: 29.9 - (2 × 2.31) = 29.9 - 4.62 = 25.3 degrees Celsius.Calculate the upper boundary by adding two standard deviations to the mean: 29.9 + (2 × 2.31) = 29.9 + 4.62 = 34.5 degrees Celsius.Thus, 95% of the time, the maximum monthly temperature is between 25.3 and 34.5 degrees Celsius.
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the differences between right and left arm blood pressure measurements. Right Arm 102 101 94 79 79 Left Arm 175 169 182 146 144 a. Find the values of d and sd (you may use a calculator).b. Construct a 90% confidence interval for the mean difference between all right and left arm blood pressure measurements.
Answer:
a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
b) The 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
Step-by-step explanation:
1) Previous concepts and notation
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Let put some notation
x=left arm , y = right arm
x: 175 169 182 146 144
y: 102 101 94 79 79
The first step is define the difference [tex]d_i=x_i-y_i[/tex], that is given so we have:
d: 73, 68, 88, 67, 65
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]
2) Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=5-1=4[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]
Now we have everything in order to replace into formula (1):
[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.330[/tex]
[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.070[/tex]
So on this case the 90% confidence interval would be given by (63.330;81.070)
[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]
Prove that two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent.
Two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent.
Explanation:In order to prove that two right triangles are congruent if the corresponding altitudes and angle bisectors through the right angles are congruent, we can use the side-side-side (SSS) congruence theorem. This theorem states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.
In this case, we can show that the corresponding altitudes and angle bisectors through the right angles are congruent for both triangles. Since both triangles have congruent corresponding altitudes and congruent angle bisectors through the right angles, we can conclude that the triangles are congruent.
Therefore, the statement is proven.
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State the half-angle identities used to integrate sin^(2) x and cos^(2) x
the half-angle formulas are sin ^(2)x = ?? and cos ^(2)x = ??
Answer:
the answer is D
Step-by-step explanation:
Think of the Pythagorean Theorem which states that a^2 + b^2 = c^2. The Pythagorean Identities used in trigonometry are the angle version which can be used to simplify expressions.
HOPE THIS HELPED ;3
The formula of half angle identities is sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
What is angle?An angle is the formed when two straight lines meet at one point, it is denoted by θ.
The given terms are,
sin²x and cos²x
To integrate sin²x, the formula is used,
sin²x = (1-cos(x/2))/2
To integrate cos²x, the formula is used,
cos²x = (1+cos(x/2))/2
The formula is, sin²x = (1-cos(x/2))/2 and cos²x = (1+cos(x/2))/2.
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A new drug test needs to be evaluated. The probability of a random person taking drugs is 4%. The drug test tested positive for a person not taking drugs 2% of the time. The drug test tested negative for a person taking drugs 1% of the time. a. What is the probability that a random person tests negative? b. What is the probability that a person who tests negative is without any drug problems?
Answer:
Step-by-step explanation:
Given that a new drug test needs to be evaluated. The probability of a random person taking drugs is 4%.
The drug test tested positive for a person not taking drugs 2% of the time. The drug test tested negative for a person taking drugs 1% of the time.
Tested positive Tested negative Total
Having drug 0.98 0.02 1.00
not hav drug 0.01 0.99 1.00
a) the probability that a random person tests negative
= Prob ( really negative and test negative) + Prob (positive and test negative)
[tex]= 0.96*0.99+0.04*0.02= 0.9504+0.0008=0.9512[/tex]
b) the probability that a person who tests negative is without any drug problems
= [tex]\frac{0.9504}{0.9504+0.0008} \\=0.9992[/tex]
For a given piece of code, the hit ratio of first cache is 0.1 and the hit ratio of second cache is 0.3. The time required to access the first cache is 10 nanoseconds, the second cache is 100 nanoseconds, and the time to access the underlying physical memory is 1 microsecond, what is the effective memory access time for the piece of code?
Answer:
effective memory access = 658 ns
Step-by-step explanation:
GIven data:
Effective memory access time is given as
[tex] = [H_1*T_1]+[(1-H_1)*H_2*T_2]+[(1-H_1)(1-H_2)*H_m*T_m][/tex]
from the data given above we have
[tex]H_1 = 0.1[/tex]
[tex]H_2 = 0.3[/tex]
[tex]T_1 = 10 ns[/tex]
[tex]T_2 = 100 ns[/tex]
hit rate, [tex]H_m = 1 ns[/tex]
access time [tex]= T_m = 1000 ns[/tex]
Plugging all information in above formula to get the effective memory access
[tex]= 0.1\times 10 + 0.9\times 100+ 0.9 \times 0.7\times 1 \times 1000[/tex]
= 1+27+ 630
=658 ns