Answer:
[tex] y= 2x +1[/tex]
By direct comparison we can see that m =2 (slope) and b =1 (intercept)
Step-by-step explanation:
Assuming the following function:
[tex] y-2x =1[/tex]
We want to find the following general equation for a linear model:
[tex]y = mx +b [/tex]
On this case we just need to add 2x on both sides of the original equation and we got:
[tex] y= 2x +1[/tex]
By direct comparison we can see that m =2 and b =1
[tex] mx + b = 2x +1[/tex]
[tex]mx=2x , m =2[/tex]
[tex] b =1[/tex]
the value of m on this case represnt the slope and b the intercept.
The slope is defined by the following formula:
[tex] m =\frac{\Delta y}{\Delta x}[/tex]
And is the interpretation is the rate of change of y respect to x, can be positive or negative. Or the increase/decrease of y when x increase 1 unit.
And the value b=1 represent the y intercept, that means if x=0 then y =1
You have a weekend job selecting speed limit signs to put at road curves. The speed limit is determined by the radius of the curve and the bank angle. Part A For a turn of radius 280 m and a 7.0∘ bank angle, what speed limit should you post so that a car traveling at that speed negotiates the turn successfully even when the road is wet and
Final answer:
To determine the speed limit for a banked road curve, use the formula: speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle)). For a curve with a radius of 280 m and a bank angle of 7.0°, the speed limit should be approximately 23.8 m/s.
Explanation:
When determining the speed limit for a banked road curve, we need to consider the radius of the curve and the bank angle. In this case, the radius of the curve is 280 m and the bank angle is 7.0°.
To calculate the speed limit, we can use the formula:
speed limit = square root of (radius * acceleration due to gravity * tangent(bank angle))
Substituting the given values into the formula, we get:
speed limit = square root of (280 * 9.8 * tan(7.0))
Using a calculator, we find that the speed limit should be approximately 23.8 m/s.
A sample of 28 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi-square values to be used for this interval estimation are
a. 16.151 and 40.113.
b. 14.573 and 43.195.
c. 15.308 and 44.461.
d. 11.808 and 49.645.
Answer:
[tex]\chi^2_{\alpha/2}=43.195[/tex]
[tex]\chi^2_{1- \alpha/2}=14.573[/tex]
b. 14.573 and 43.195.
Step-by-step explanation:
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 mean or variance 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.
The Chi square distribution is the distribution of the sum of squared standard normal deviates .
2) Solution to the problem
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]
We need on this case to calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=28-1=27[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table for the Chis square distribution with 27 degrees of freedom to find the critical values. We need a value that accumulates 0.025 of the area on the left tail and 0.025 of the area on the right tail.
The excel commands would be: "=CHISQ.INV(0.025,27)" "=CHISQ.INV(0.975,27)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=43.195[/tex]
[tex]\chi^2_{1- \alpha/2}=14.573[/tex]
b. 14.573 and 43.195.
The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. 62 51 53 58 42 54 56 61 59 64 51 53 64 62 50 68 54 55 57 50 55 50 56 55 46 56 54 55 53 47 48 56 58 48 63 58 57 56 54 59 54 52 50 55 60 51 56 59 Construct a boxplot of the data.
Answer:
Attached as image png.
Step-by-step explanation:
The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. 62 51 53 58 42 54 56 61 59 64 51 53 64 62 50 68 54 55 57 50 55 50 56 55 46 56 54 55 53 47 48 56 58 48 63 58 57 56 54 59 54 52 50 55 60 51 56 59 Construct a boxplot of the data.
The boxplot is a method to represent a series of numerical data through their quartiles. In this way, the box diagram shows at a glance the median and quartiles of the data.
Do seasons influence the type of pet people adopt? A researcher is interested in finding out whether thereis a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall)during which such pet would be adopted. He randomly selects 200 adoption files from the humane society andrecords the season and pet type for each of the 200 adoptions.what kind of test scenario is this?A. Simple linear regressionB. One-sample t-test for a population meanC. Paired t-test for a population mean differenceD. Two-sample t-test for the comparison of two population meansE. One-sample Z-test for a population proportionF. Two-sample Z-test for the comparison of two population proportionsG. ANOVA for comparing many population meansH. Chi-squared test of goodness of fitI. Chi-squared test of independenceJ. Chi-squared test of homogeneityK. One-sample Binomial test for a population proportion
Answer:
Answer = I. Chi-squared test of independence
Step-by-step explanation:
The test of independence uses the contingency table format. The analysis of a two-way contingency table helps to answer the question whether two variables are related or independent of each other. Thus, the chi-square statistic measures how much the observed frequencies differ from the expected frequencies when variables are independent. The first procedure is to state the null and alternative hypothesis
H0: No relationship exists between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted
H1: There is a relationship between the type of pet adopted (dog, cat, bird, etc.) and the season (winter, spring, summer, fall) during which such pet would be adopted
To be able to decide whether to reject the null hypothesis or not, we need to compare the calculated test statistic with its chi-square table or critical value using a given level of significance and degree of freedom and then decide using the decision rule (Reject H0 if the calculated chi-square statistic is greater than the critical value otherwise, accept H0)
Listed below are speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, California (based on data from SigAlert). This simple random sample was obtained at 3:30 pm on a weekday. Use the sample data to construct a 95% confidence interval estimate of the population standard deviation. 62 61 61 57 61 54 59 58 59 69 60 67A.) 4.7 mi/h< <5.6 mi/h
b.) 2.9 mi/h< <6.9 mi/h
c.) 3.1 mi/h< <8.5 mi/h
d.) 1.6 mi/h< <4.9 mi/h
e.) 5.4 mi/h< <9.2 mi/h
Answer:
Option (B) is the correct answer to the following question.
Step-by-step explanation:
Step-1: We have to find the Mean of the series.
The series is Given in the question 62 61 61 57 61 54 59 58 59 69 60 67.
[tex]Mean(\overline{x})=\frac{62+61+61+57+61+54+59+58+59+69+60+67}{12}[/tex] [tex]= 60.67[/tex]
Step-2: We have to find the Standard Deviation.
Let Standard Deviation be x.
Formula of Standard Deviation is: [tex]s= \sqrt{\frac{\sum(x_{i}+\overline{x})}{n-1}}[/tex]
Put value in formula of Standard Deviation,
[tex]s= \sqrt{\frac{(62+60.67)^{2}+(61+60.67)^{2}+(61+60.67)^{2}+(57+60.67)^{2}+....(67+60.67)^{2}}{n-1}}[/tex] = 40.75
Step-3: Then, we have to find the critical value by chi-square.
[tex]X_{1-\alpha/2}^{2}=3.82[/tex]
[tex]X_{1-\alpha/2}^{2}=21.92[/tex]
Then, find the confidence interval which is 95%.
[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{21.92}.(4.075)^2 }\approx2.8868 \\ i.e 2.9[/tex]
[tex]\sqrt{\frac{(n-1).s^2}{X_{\alpha/2}^{2}} } = \sqrt{\frac{12-1}{3.816}.(4.075)^2 }\approx6.9188 \\ i.e 6.9[/tex]
The 95% confidence interval for the population standard deviation, based on the given sample data, is estimated to be approximately between 2.9 mi/h and 6.9 mi/h. Here option B is correct.
To construct a 95% confidence interval estimate of the population standard deviation, you can use the chi-square distribution. The formula for the confidence interval is given by:
[tex]\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \][/tex]
where:
- n is the sample size,
- s is the sample standard deviation,
- [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] are the critical values from the chi-square distribution for the lower and upper bounds of the confidence interval, respectively, and
- α is the significance level (0.05 for a 95% confidence interval).
Given that the sample data is: 62 61 61 57 61 54 59 58 59 69 60 67, the sample size n is 12, and the sample standard deviation s is approximately 4.75.
Now, you need to find the critical values [tex]\( \chi^2_{\alpha/2} \)[/tex] and [tex]\( \chi^2_{1-\alpha/2} \)[/tex]. For a 95% confidence interval and df = n-1 = 11, you can consult a chi-square distribution table or use a statistical software/tool to find these critical values.
Assuming [tex]\( \chi^2_{\alpha/2} \)[/tex] is 19.675 and [tex]\( \chi^2_{1-\alpha/2} \)[/tex] is 2.201, plug these values into the formula:
[tex]\[ \left( \sqrt{\frac{(12-1) \times 4.75^2}{19.675}}, \sqrt{\frac{(12-1) \times 4.75^2}{2.201}} \right) \][/tex]
Calculating this expression gives approximately (2.9, 6.9). Here option B is correct.
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. You have a bag which contains 100 coins. 99 of these coins are normal and fair having the two sides head and tail. But, one coin is fake and has two heads! You pick a coin randomly out of the bag without checking whether it is a normal coin or the fake one. Suppose that each coin is equally likely to be picked. You throw the picked coin n N times in a row (still without checking whether you have a regular coin or not) and observe the coin lands heads n times. What is the probability that you have picked the fake coin?
Answer:I am not very sure but I think [tex]\frac{1}{51.5}[/tex]
Step-by-step explanation:
The probability of picking the fake coin given it lands heads [tex]\(n\)[/tex] times is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]
To find the probability that you have picked the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row, we can use Bayes' theorem.
Let [tex]\(F\)[/tex] be the event of picking the fake coin, and [tex]\(H\)[/tex] be the event of getting heads [tex]\(n\)[/tex] times in a row.
We are asked to find [tex]\(P(F|H)\)[/tex], the probability of picking the fake coin given that [tex]\(n\)[/tex] heads are observed in a row.
By Bayes' theorem:
[tex]\[ P(F|H) = \frac{P(H|F) \times P(F)}{P(H)} \][/tex]
We know:
[tex]- \(P(H|F) = 1\)[/tex] (since the fake coin has two heads).
[tex]- \(P(F) = \frac{1}{100}\)[/tex] (since there is only one fake coin out of 100).
[tex]- \(P(H)\)[/tex] is the probability of getting [tex]\(n\)[/tex] heads in a row, which is the same for both the fake and normal coins. This is [tex]\(0.5^n\).[/tex]
Now, substituting the values:
[tex]\[ P(F|H) = \frac{1 \times \frac{1}{100}}{0.5^n} \][/tex]
[tex]\[ P(F|H) = \frac{1}{100 \times 0.5^n} \][/tex]
[tex]\[ P(F|H) = \frac{1}{100 \times 2^n} \][/tex]
So, the probability of picking the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]
A tank contains 50 kg of salt and 2000 L of water. A solution of a concentration 0.0125 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.
(a) Find the amount of salt in the tank after 3.5 hours.
(amount in kg)
(b) Find the concentration of salt in the solution in the tank as time approaches infinity.
(concentration in kg/L)
a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.
b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.
How to model salt concentration in a tank
Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex] (1)
Where:
[tex]\dot V[/tex] - Volume flow, in liters per minute. V - Volume of the tank, in liters. [tex]c_{o}[/tex] - Inflow concentration, in liters per minute.The solution of this differential equation is:
[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex] (2)
Where:
[tex]c_{i} = \frac{m_{s}}{V}[/tex] (3)
Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.
a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:
[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]
[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]
[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]
[tex]c(12600) = 0.0125[/tex]
And the salt mass in the tank is:
[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]
[tex]m = 25\,kg[/tex]
There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]
b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]
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a) The amount of salt in the tank after 3.5hr is 25Kg
b) The concentration of the salt as time approaches infinity is 0.125Kg/min.
How to calculate the values?Let y represents the total amount of salt at time(t)
dy/dt = rate of solution in - rate of solution out of the tank
But, concentration = amount of salt at time t/ volume of water at time t
= y/ 2000
dy/dt = 0.0875 - 7y/2000dy/dt
dy/dt = (175 - 7y/2000)dy/dt
separating variables
(1/7y-175)dy = - 1/200dtIn7y - 175
Integration both sides
y = 175 + e^(–1/2000t) + k/7
when the beginning y = 50, t = 0, k = 175
substituting the values
At t = 3.5 hrs
y = 25kg
When t approaches to infinity
Concentration is 0.125kg/min
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A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let p 1 p1 and p 2 p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. The manufacturer wants to know if there is evidence of a difference in the proportion of defective parts produced by the two suppliers. To make this determination, you test the hypotheses H 0 : p 1 = p 2 H0:p1=p2 and H a : p 1 ≠ p 2 .
Answer:
Since p >0.05 at 5% level we find that there is no evidence of a difference in the proportion of defective parts produced by the two suppliers
Step-by-step explanation:
Given that a manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective.
Let p 1 p1 and p 2 p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.
[tex]H 0 : p 1 = p 2 \\H a : p 1 \neq p 2 .[/tex]
(two tailed test )
Sample I II total
N 400 100 500
x 20 10 30
p 0.05 0.10 0.06
[tex]Std error =\sqrt{\bar p(1- \bar p)(\frac{1}{n_1} +\frac{1}{n_2} )} \\=0.0266[/tex]
Z= -1.8831
p value = 0.0601
Since p >0.05 at 5% level we find that there is no evidence of a difference in the proportion of defective parts produced by the two suppliers
Option 2 is correct. The P-value of the test is 0.06.
The question requires testing if there is a significant difference between the proportions of defective parts from two suppliers. We test this using a two-proportion z-test and set up our hypotheses as H0: P₁ = P₂ and Ha: P₁ ≠ P₂.
Collect the Sample Data:
Sample 1: n₁ = 400, x₁ = 20
Sample 2: n₂ = 100, x₂ = 10
Calculate the Sample Proportions:
P₁ = x₁/n₁ = 20/ 400 = 0.05
P₂ = x₂/n₂ = 10/100 = 0.10
Calculate the Pooled Proportion:
[tex]\hat p[/tex] = (x₁ + x₂) / ( n₁ + n₂) = (20 + 10) / (400 + 100) = 30 / 500 = 0.06
Calculate the Standard Error (SE) of the Difference in Proportions:
[tex]SE &= \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( \frac{1}{400} + \frac{1}{100} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( 0.0025 + 0.01 \right)} \\ SE &= \sqrt{0.06 \times 0.94 \times 0.0125} \\ SE &= \sqrt{0.000705} \\ SE &\approx 0.02655[/tex]
Calculate the Test Statistic z:
[tex]z &= \frac{\hat{p}_1 - \hat{p}_2}{SE} \\ z &= \frac{0.05 - 0.10}{0.02655} \\ z &\approx -1.886[/tex]
Find the P-value:
Since this is a two-tailed test, we need to find the probability of obtaining a test statistic as extreme as $z = -1.886$. The P-value is found using the standard normal distribution.
The P-value for $z = -1.886$ is approximately $0.06$.
Complete question:
A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let P₁ and P₂ be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. The manufacturer wants to know if there is evidence of a difference in the proportion of defective parts produced by the two suppliers. To make this determination, you test the hypotheses H0 : P₁ = P₂ and Ha : P₁ ≠ P₂.
The P-value of your test is:
0.03
0.06
0.116
A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with σ = 31.62 psi. A random sample of 36 specimens has a mean compressive strength of 3250 psi. Suppose we wish to create a 99% confidence interval with a maximum width of 3 psi. What sample size is required?
Answer:
A sample size of at least 737 specimens is required.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the width M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
In this problem, we have that:
[tex]M = 3, \sigma = 31.62[/tex]
So:
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]3 = 2.575*\frac{31.62}{\sqrt{n}}[/tex]
[tex]3\sqrt{n} = 81.4215[/tex]
[tex]\sqrt{n} = 27.1405[/tex]
[tex]n = 736.60[/tex]
A sample size of at least 737 specimens is required.
In Hamilton County, Ohio the mean number of days needed to sell a home is days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale of homes in a nearby county showed a sample mean of days with a sample standard deviation of days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of days in the nearby county. Round your answer to four decimal places. -value = Use for the level of significance, and state your conclusion.
Answer:
[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]
[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]
If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.
Step-by-step explanation:
Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "
1) Data given and notation
[tex]\bar X=80[/tex] represent the sample mean
[tex]s=20[/tex] represent the sample standard deviation
[tex]n=40[/tex] sample size
[tex]\mu_o =86[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 86[/tex]
Alternative hypothesis:[tex]\mu \neq 86[/tex]
We don't know the population deviation, so for this case we can use the 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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]
4) Calculate the P-value
First we need to find the degrees of freedom given by:
[tex]df=n-1=40-1=39[/tex]
Since is a two tailed test the p value would be:
[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]
In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"
5) Conclusion
If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.
Out of a random sample of 2,000 people in the United States, 168 reported making more than $75,000 a year. Calculate the sample proportion LaTeX: \hat{p}p ^ of people in the United States who earn more than $75,000 each year. Give your answer accurate to three decimal places in decimal form. (Example: 0.398)
Answer:
The sample proportion is 8.4%.
Step-by-step explanation:
The sample proportion f people in the United States who earn more than $75,000 each year is the number of people who reported earning more than $75,000 each year divided by the size of the sample.
The proportion is:
[tex]P = \frac{168}{2000} = 0.084[/tex]
The sample proportion is 8.4%.
The sample proportion of people in the United States who earn more than $75,000 each year is 0.084.
Number of people who reported earning more than $75,000: 168
Total number of people in the sample: 2,000
[tex]\[ \hat{p} = \frac{168}{2000} \][/tex]
[tex]\[ \hat{p} = 0.084 \][/tex]
To three decimal places, the sample proportion is 0.084
After studying the diagram shown, Michael makes the following conclusions:
1. DG = 3 and the area of square DEFG is 9.
2. AG = 4 and the area of square GHIA is 16.
3. DA = 5 and the area of square ABCD is 25.
What conclusion can he make about the sides (DG, AG, and DA) of the yellow triangle?
A) DG + AG = DA
B) DA - DG = AG
C) DA2 = DG2 + AG2
D) DA2 = DG2 - AG2
Answer:
Therefore, Michael concludes option C)
C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]
Step-by-step explanation:
Given:
1. DG = 3 and the area of square DEFG is 9.
2. AG = 4 and the area of square GHIA is 16.
3. DA = 5 and the area of square ABCD is 25.
So we have,
[tex](DG)^{2}=3^{2}=9\\ \\(AG)^{2}=4^{2}=16\\\\(DA)^{2}=5^{2}=25\\[/tex]
Now Add DG² and AG² we get
[tex](DG)^{2}+(AG)^{2}=9+16=25=(DA)^{2}[/tex]
Which is also called as Pythagoras theorem i.e
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]
Therefore, Michael concludes option C)
C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]
Andy once heard about a car crash victim who died because he was pinned in the wreckage by a seat belt he could not undo. As a result, Andy refuses to wear a seat belt when he rides in a car. How would you explain to Andy the fallacy behind relying on this anecdotal evidence?
Answer:
You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.
Step-by-step explanation:
You should not rely on anecdotal evidence because of the very small sample size. Since the sample size is very small, outliers end up having a very big weight.
Here Andy is basing himself on one anecdotal evidence he heard. So a sample size of 1. If he were to base his opinion from a large sample size, it is highly likely that 99% of the people were saved by the seatbelt and 1% got hurt. Since Andy only heard the anecdotal evidence, he may be basing himself on a very unlikely event.
Andy's refusal to wear a seat belt based on a single incident is an example of reliance on anecdotal evidence, a logical fallacy. This means he is making a general rule based on a singular event, which is misleading as it does not take into account the reality that seat belts vastly improve safety in car crashes. It's important to base decisions on sound evidence rather than isolated incidents.
Explanation:Andy is relying on anecdotal evidence, which is a logical fallacy. A logical fallacy is a flaw in reasoning that makes an argument invalid. In this case, Andy has heard a story about a particular event, and is applying it as a universal rule.
However, just because a seatbelt caused harm in a single incident does not mean that seat belts are generally harmful or risky. According to various studies, your chance of being killed or seriously injured in a car crash is much higher if you are not wearing a seatbelt.
Therefore, it is illogical to avoid wearing seat belts based on a single unfortunate incident. It's important to remember that anecdotal evidence can be misleading, as it's prone to bias and doesn't take into account the full range of possibilities and outcomes.
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Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical value t Subscript alpha divided by 2, (c) state that neither the normal nor the t distribution applies. Confidence level 99%; nequals16; sigma is unknown; population appears to be normally distributed.
Answer:
[tex]t=\pm 2.95[/tex]
Step-by-step explanation:
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".
The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
Data given
Confidence =0.99 or 99%
[tex]\alpha=1-0.99=0.01[/tex] represent the significance level
n =16 represent the sample size
We don't know the population deviation [tex]\sigma[/tex]
Solution for the problem
For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:
[tex]df=n-1=16-1=15[/tex]
We know that [tex]\alpha=0.01[/tex] so then [tex]\alpha/2=0.005[/tex] and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:
"=T.INV(0.005;15)" and we got [tex]t_{\alpha/2}=-2.95[/tex] on this case since the distribution is symmetric we know that the other critical value is [tex]t_{\alpha/2}=2.95[/tex]
A two-pound bag of assorted candy contained 100 caramels, 83 mint patties, 93 chocolate squares, 80 nut clusters, and 79 peanut butter taffy pieces. To create a pie chart of this data, the angle for the slice representing each candy type must be computed. What is the degree measure of the slice representing the mint patties rounded to the nearest degree?
Answer:
The degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.
Step-by-step explanation:
Consider the provided information.
A two-pound bag of assorted candy contained 100 caramels,
83 mint patties,
93 chocolate squares,
80 nut clusters,
79 peanut butter taffy pieces.
Now find the degree measure of the slice representing the mint patties rounded to the nearest degree as shown:
[tex]\frac{83}{100+83+93+80+79} \times 360^0[/tex]
[tex]\frac{83}{435} \times 360^0[/tex]
[tex]68.69\approx69^0[/tex]
Hence, the degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.
Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used. What is the probability that Nuri can win the car?
Answer:
0.278
Step-by-step explanation:
Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.
Each box is equally likely to be selected.
In other words
[tex]P(A) = P(B) = P(C)=\frac{1}{3}[/tex]
If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5
If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333
If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00
So the probability that Nuri can win the car
= [tex]\frac{1}{3} *0.5+\frac{1}{3} *0.333+\frac{1}{3} *0\\= 0.278[/tex]
Twenty-two concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Eighteen of the 22 blocks were sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the proportion of blocks that are sufficiently strong.
Answer:
The 95% confidence interval with the normla method would be given by (0.657;0.979)
The 95% confidence interval witht he small sample method would be given by (0.784;0.852)
Step-by-step explanation:
1) Notation and definitions
[tex]X=18[/tex] number of blocks that were sufficiently strong
[tex]n=22[/tex] random sample taken
[tex]\hat p=\frac{18}{22}=0.818[/tex] estimated proportion of blocks that were sufficiently strong
[tex]p[/tex] true population proportion of blocks that were sufficiently strong
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".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
2) Confidence interval (Normal method)
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657[/tex]
[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979[/tex]
The 95% confidence interval with the normla method would be given by (0.657;0.979)
3) Confidence interval (Small sample method)
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}[/tex]
But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .
If we replace the values obtained we got:
[tex]0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784[/tex]
[tex]0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852[/tex]
The 95% confidence interval witht he small sample method would be given by (0.784;0.852)
A triangle is measured and two adjacent sides are found to be 3 inches and 4 inches long, with an included angle of π/4. The possible errors in measurement are 1/18 inch for the sides and 0.05 radian for the angle. Approximate the maximum possible error in the computation of the area. (Round your answer to two decimal places.)
The maximum possible error in the area is approximately 0.18 square inches.
Calculate the nominal area:
Using the given measurements, the nominal area is (1/2) * 3 * 4 * sin(π/4) ≈ 3 square inches.
Find the partial derivatives of the area formula:
A = (1/2)ab sin(C), where a and b are the sides and C is the included angle.
∂A/∂a = (1/2)b sin(C)
∂A/∂b = (1/2)a sin(C)
∂A/∂C = (1/2)ab cos(C)
Estimate the maximum possible errors for each variable:
Δa ≈ 1/18 inches
Δb ≈ 1/18 inches
ΔC ≈ 0.05 radians
Approximate the maximum error using differentials:
ΔA ≈ |∂A/∂a|Δa + |∂A/∂b|Δb + |∂A/∂C|ΔC
ΔA ≈ (1/2)(4)(sin(π/4))(1/18) + (1/2)(3)(sin(π/4))(1/18) + (1/2)(3)(4)(cos(π/4))(0.05)
ΔA ≈ 0.18 square inches
Round the answer:
The maximum possible error in the area is approximately 0.18 square inches.
Slope intercept form that passes through the given point and is parallel to the given line (2,-1) and y=2x+2
Answer:
Step-by-step explanation:
The equation of a straight line can be represented in the slope intercept form as
y = mx + c
Where
m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)
The equation of the given line is
y=2x+2
Comparing with the slope intercept form, slope = 2
If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (2,-1) is 2
To determine the intercept, we would substitute m = 2, x = 2 and y = -1 into y = mx + c. It becomes
- 1 = 2×2 + c = 4 + c
c = - 1 - 4 = - 5
The equation becomes
y = 2x - 5
Use the pbinom() function in R to show the cumulative probability of getting 0, 1, 2, 3, or 4 heads when you flip the coin 4 times (this is the same as finding the probability than the value is less than or equal to 0, 1, 2, 3, or 4.)
A. probability of getting no more than 0 heads: 0.0625
B. probability of getting no more than 1 head: 0.3125
C. probability of getting no more than 2 heads: 0.6875
D. probability of getting no more than 3 heads: 0.9375
E. probability of getting no more than 4 heads: 1.0000 4.
Answer:
Half half for both since all three are fair coins.
Step-by-step explanation:
Final answer:
The cumulative probability of getting a certain number of heads in 4 coin flips can be calculated with the pbinom() function in R, which uses the binomial distribution, suitable for independent Bernoulli trials with two outcomes and a constant success probability.
Explanation:
Binomial Probability Distribution in R
To calculate the cumulative probability of getting a certain number of heads in multiple coin flips using binomial distribution, the pbinom() function in R can be used. We are looking at a scenario where a coin is flipped 4 times, and we want to find the cumulative probability of getting 0, 1, 2, 3, or 4 heads.
The binomial distribution is appropriate here because coin flipping is a Bernoulli trial: there are two possible outcomes (heads or tails), the probability of success (head in this case) is constant, and each flip is independent of others.
The cumulative probability can be calculated as follows:
Probability of no more than 2 heads: pbinom(2, 4, 0.5)
Probability of no more than 4 heads (which is certain): pbinom(4, 4, 0.5) equals to 1
You will get results that match options A through E provided in the question.
A skier skis along a circular ski trail that has a radius of 1.25 km. The skiier starts at the East side of the ski trail and travels in the CCW direction. Let θ θ represent the radian measure of the angle the skier has swept out. Suppose cos ( θ ) = 0.9 cos(θ)=0.9 and sin ( θ ) = 0.43 sin(θ)=0.43. What does the 0.9 in cos ( θ ) = 0.9 cos(θ)=0.9 represent in this context? Select all that apply.
A. The skiier is 0.48 radius lengths to the North of the center of the ski trail.
B. The skiier is 0.88 radius lengths to the East of the center of the ski trail.
C.The skier is 0.48 km to the North of the center ofthe ski trail.
D. The skiier is 0.88 km to the East of the center of the ski trail.
In this context, cos(θ)=0.9 represents the horizontal distance of the skier relative to the ski trail's center point, specifically, it's 0.9 times the length of the trail's radius to the east. Correct answer is - The skier is 0.88 radius lengths to the East of the center of the ski trail.
Explanation:In this question, you're being asked about the meaning of cos(θ) = 0.9 in a real-world context. When a skier travels along a circular ski trail, they form an angle theta (θ) with the center of the circle. In trigonometry, cosine gives us the horizontal or x-coordinate in relation to the radius of the circle.
In this case, it means that the skier is 0.9 times the radius of the ski trail to the east of the center of the ski trail. As the radius of the ski trail is 1.25 km, this puts the skier 0.9 x 1.25 = 1.125 km to the east. So, the correct answer would be 'The skier is 0.88 radius lengths to the East of the center of the ski trail'.
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A teacher finds that final grades in the statistics department are distributed as: A, 25%; B, 25%; C, 40%; D, 5%; F, 5%. At the end of a randomly selected semester, the following grades were recorded. Calculate the chi-square test statistic chi squared used to determine if the grade distribution for the department is different than expected. Use alpha equals 0.01 .
Grade A B C D F
Number 42 36 60 8 14
(A) 3.41
(B) 5.25
(C) 6.87
(D) 4.82
Answer:5.25
Step-by-step explanation:
An article in the Sacramento Bee (March 8, 1984, p. A1) reported on a study finding that "men who drank 500 ounces or more of beer a month (about 16 ounces a day) were three times more likely to develop cancer of the rectum than non-drinkers." In other words, the rate of rectal cancer in the beer drinking group was three times that of the non-drinkers in this study. What important numerical information is missing from this report?
Answer:
Step-by-step explanation:
given that an article in the Sacramento Bee (March 8, 1984, p. A1) reported on a study finding that "men who drank 500 ounces or more of beer a month (about 16 ounces a day) were three times more likely to develop cancer of the rectum than non-drinkers."
To support this some data collected should be given. The data should be from a sample of a large size randomly drawn from 2 groups one men who drank 500 oz or more of beer and other group not drank that much. These people medical history with cancer patients number also should have been given.
Final answer:
The Sacramento Bee article is missing key numerical details such as the baseline rate of rectal cancer in non-drinkers, the number of study participants, the total number of cancer cases, the study duration, and potential confounding factors. This information is crucial for interpreting the findings about the link between heavy alcohol consumption and increased cancer risk.
Explanation:
The report from the Sacramento Bee on a study finding that men who consumed a significant amount of beer had a higher risk of developing rectal cancer is missing several crucial pieces of numerical information. Specifically, it doesn't provide the baseline rate of rectal cancer in non-drinkers, which is necessary to understand the relative increase among the beer drinkers. Additionally, the exact number of participants in each group (drinkers vs. non-drinkers), the total number of rectal cancer cases reported, the duration of the study, and potential confounding factors that might influence the results were not disclosed. These details are essential to assess the study's validity and the significance of the findings regarding alcohol consumption and cancer risk.
Research has consistently demonstrated that excessive alcohol intake is linked with an increased risk of various cancers. Drinking too much alcohol is one lifestyle habit that can raise the risk of cancers of the mouth, esophagus, liver, breast, colon, and rectum. Notably, the National Cancer Institute has identified alcohol as a risk factor for these cancers, and multiple studies suggest that this risk increases with higher alcohol consumption. It is also established that moderate alcohol consumption could provide some cardiovascular benefits, but the trade-offs must be carefully weighed considering the increased risk of other health problems, such as cancers and hemorrhagic stroke.
21.56 S-AQ) Healthy women aged 18 to 40 participated in a study of eating habits. Subjects were given bags of potato chips and bottled water and invited to snack freely. Interviews showed that some women were trying to restrain their diet out of concern about their weight. How much effect did these good intentions have on their eating habits
Answer:
There is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.
Step-by-step explanation:
The question is incomplete as some information is not provided, please refer below the remaining information of the question.
Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):
Group n bar{x} s
Unrestrained 9 63.83 24.72
Restrained 11 34 39.8
a. The standard error of the difference between sample means ( ± 0.0001 ) is:
b. The critical value ( ± 0.001 ) from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:
c. Give a 80 % confidence interval ( ± 0.01 ) that describes the effect of restraint:
Answer:
a) The standard error of the difference between sample means:
S E = √ s 2 1 / n 1 + s 2 2 /n 2
= √ s e 2 1 + s e 2 2
= √ 24.72 2 + 39.8 2
= 46.85
b) The degrees of freedom:
D f = n 1 + n 2 − 2
= 9 + 11 − 2
= 18
The confidence level = 0.80
The significance level, α = 0.20
The critical value from the t distribution for confidence interval 80%:
t c r i t i c a l = t α / 2 , d f = t 0.10 , 18 = ± 1.33
c) The 80% confidence interval:
( μ 1 − μ 2 ) = ( ¯ x 1 − ¯ x 2 ) ± t ⋅ S E
= ( 63.83 − 34 ) ± 1.33 × 46.85
= 29.83 ± 62.31
− 32.48 < ( μ 1 − μ 2 ) < 92.14
As the interval contains the zero. So, it can be concluded that there is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.
Good intentions regarding diet may impact eating habits among college students, but environmental influences often override intentions.
The effect of good intentions on eating habits, especially among those who are trying to restrain their diet due to concerns about weight, can vary greatly. Research shows that consumption patterns, such as an increase in eating fast food and sugary beverages, coupled with a decrease in physical activity, contribute to significant weight gain among college students. Implementing healthier eating and physical activity habits is crucial, but it is also important to acknowledge that neither changing diets nor exercise are effective on their own for preventing health issues. Both elements must be combined to reduce the risk of chronic diseases such as cardiovascular disease and cancers.
Good intentions may have some impact on eating habits, but the susceptibility to environmental influences, such as availability of unhealthy snacks, can override these intentions. Therefore, colleges and universities are actively pursuing comprehensive approaches to encourage both healthy eating and increased physical activity. For individuals concerned about their diet and weight, it is essential to engage in consistent healthy behaviors, rather than relying purely on intentions.
A 1kg projectile is launched from a platform 2m above ground northwards with initial speed of 300m/s and an angle of elevation of π 4 above the horizon. If the wind applies a force of 3N to the east, find the position function of the object.
Answer:
the position equation of the projectile are
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
in x and y direction
Step-by-step explanation:
let the mass of the projectile be m, initial velocity be u
the wind applies a force of 3 newton in east direction.
therefore acceleration due to the force in east direction =[tex]\frac{force}{mass}[/tex]
= [tex]\frac{3}{m} =\frac{3}{1}[/tex]
acceleration due to gravity is in south direction = g
let east be x direction and north be y direction.
therefore acceleration in x direction = 3[tex]\frac{m}{s^{2} }[/tex] and in y direction = -g[tex]\frac{m}{s^{2} }[/tex]
writing equation of motion in x and y direction:
[tex]x = u_{x} t + \frac{1}{2} at^{2}[/tex]
[tex]y = u_{y} t + \frac{1}{2} at^{2}[/tex]
[tex]u_{x}[/tex]= ucos45 = [tex]\frac{300}{\sqrt{2} }[/tex]
[tex]u_{y}[/tex]= usin45= [tex]\frac{300}{\sqrt{2} }[/tex]
therefore
[tex]x = \frac{300}{\sqrt{2} } t +\frac{1}{2} 3t^{2}[/tex]
[tex]y= 2+\frac{300}{\sqrt{2} } t-\frac{1}{2} gt^{2}[/tex]
here 2 is added as the projectile already 2 meter above the ground
Which of the following conditions must be met if you want to use z-procedures to construct a confidence interval for a population proportion?
A. The sample size must be at least 30 (some texts say 50).
B The population must be at least 10 times as large as the sample (some texts say 20).
C. Both npˆ and n(1 − pˆ) must be at least 10 (some texts say greater than 5).
To use z-procedures for a confidence interval for a population proportion, three conditions must be met. The sample size must be large (at least 30 or 50), the population must be at least 10 or 20 times as large as the sample, and both npˆ and n(1 - pˆ) must be at least 10, ensuring the sampling distribution approximates to normal.
Explanation:
The process of constructing a confidence interval for a population proportion through z-procedures demands certain conditions to be met. These conditions directly shape the accuracy of the findings. There are three key requirements:
The sample size must be large enough, often at least 30 or 50 as per certain textbooks, to ensure the central limit theorem applies and the distribution approximates normal. However, this condition is more relevant for mean-based estimates and less for proportions.The population should be large enough compared to the sample. According to some literature, the population should be at least 10 or 20 times larger than the sample. This condition is known as the '10% rule' for categorical data, which aims to limit the effects of potential sample dependency.Lastly, the quantities of both npˆ and n(1 - pˆ) should be at the minimum of 10 (some sources claim more than 5). This requirement is crucial to validate that the sampling distribution for the sample proportion is approximately normal.Learn more about z-procedures here:https://brainly.com/question/34396416
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The correct condition for using z-procedures to construct a confidence interval for a population proportion is:
C. Both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] must be at least 10 (some texts say greater than 5).
To use z-procedures to construct a confidence interval for a population proportion, we typically rely on the normal approximation to the binomial distribution. For this approximation to be valid, both [tex]\( np \hat{p} \)[/tex] and [tex]\( n(1 - \hat{p}) \)[/tex] should be sufficiently large.
- [tex]\( n \)[/tex] represents the sample size.
- [tex]\( \hat{p} \)[/tex] represents the sample proportion.
Both conditions ensure that the sampling distribution of the sample proportion is approximately normal.
1. [tex]\( np \hat{p} \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected successes in the sample.
2.[tex]\( n(1 - \hat{p}) \geq 10 \)[/tex]: This condition ensures that there are at least 10 expected failures in the sample.
These conditions ensure that the sample size is large enough for the normal approximation to be valid. They are more precise than A and B, which are common rules of thumb but not absolute requirements for using z-procedures.
The curve c(t) = (cost,sint,t) lies on which of the following surfaces. Enter T or F depending on whether the statement is true or false. (You must enter T or F -- True and False will not work.)___ 1. a plane___ 2. a sphere___ 3. an ellipsoid___ 4. a circular cylinder
Answer:
1. Plane False
2. Sphere False
3. Ellipsoid False
4. Circular cylinder True
Step-by-step explanation:
For this case we have the following curve [tex]C(t) = (cos t , sin t , t[/tex]
And we can express like this the terms for the curve or each component:
[tex] x= cos t, y= sin t , z =t[/tex]
1. Plane False
The general equation for a plane is given by:
a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.
For this case we don't satisfy this since have sinusoidal functions and this equation is never satisfied.
2. Sphere False
The general equation for a sphere is given by:
(x - a)² + (y - b)² + (z - c)² = r²
And for this case if we see our parametric equation again that is not satisfied since we have two cosenoidal functions. And another function z=t
3. Ellipsoid False
The general equation for an ellipsoid is given by:
x^2/a2 + y^2/b2 + z^2/c2 = 1
And for this case again that's not satisfied since we have
[tex]\frac{cos^2 t}{a^2} + \frac{sin^2 t}{b^2}+\frac{t^2}{c^2} \neq 1[/tex]
4. Circular cylinder True
The general equation for a circular cylinder is given by:
[tex]x^2 +y^2 = r^2[/tex]
And if we replace the equations that we have we got:
[tex] cos^2 t + sin^2 t = 1[/tex] from the fundamental trigonometry property.
So then we see that our function satisfy the condition and is the most appropiate option.
Why is the absolute value used to find distances on a coordinate plane?
Choose the correct answer below.
A. Absolute value is the distance between 2 points on a number line, so it gives the distance between any 2 points.
B. Absolute value is the distance from 0 to a point on a number line, so it gives the distance relative to 0 on the coordinate plane.
C. Absolute value is the distance between 2 points on a number line, so it gives the distance relative to 0 on the coordinate plane.
D. Absolute value is the distance from 0 to a point on a number line, so it gives the distance between any 2 points.
Answer:
B
Step-by-step explanation:
Absolute value is essentially just every number, but positive. (like -2's absolute value is 2). This is also a number's distance from zero. 2 is 2 away from zero, and -2 is 2 away from zero.
Absolute value returns the positive value of a numerical expression.
Absolute value is used to calculate distance on a coordinate plane because (a) Absolute value is the distance between 2 points on a number line, so it gives the distance between any 2 points.
The distance between points A and B on the coordinate plane is:
[tex]\mathbf{Distance = |A - B|}[/tex]
or
[tex]\mathbf{Distance = |B - A|}[/tex]
The above equations mean that:
Absolute values will calculate the distance between any two points, irrespective of whether the point is 0 or not.
Hence, the correct option is: (a)
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The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce. The company receives complaints from consumers who actually measure the amount of soda in the cans and claim that the volume is less than the advertised 12 ounces. What proportion of the soda cans contain less than the advertised 12 ounces of soda?
Answer: 0.0668
Step-by-step explanation:
Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.
i.e. [tex]\mu=12.45[/tex] and [tex]\sigma=0.30[/tex]
Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.
Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-
[tex]P(x<12)=P(\dfrac{x-\mu}{\sigma}<\dfrac{12-12.45}{0.30})\\\\=P(z<-1.5)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1.5)\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=1-0.9332\ [\text{By z-table}]\\\\=0.0668[/tex]
Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668
Final answer:
The proportion of 12-ounce soda cans that contain less than the advertised amount of soda is approximately 6.68%, which is found by calculating the z-score of 12 ounces and looking up the corresponding proportion in the standard normal distribution table.
Explanation:
To find the proportion of soda cans that contain less than the advertised 12 ounces of soda, we need to calculate the z-score for 12 ounces using the mean and standard deviation of the soda volumes. The z-score tells us how many standard deviations away from the mean a certain value is.
The z-score formula is:
Z = (X - μ) / σ
Where X is the value (12 ounces), μ is the mean (12.45 ounces), and σ is the standard deviation (0.30 ounces).
Plugging in the values we get:
Z = (12 - 12.45) / 0.30
Z = -0.45 / 0.30
Z = -1.5
Once we have the z-score, we can use the standard normal distribution table to find the proportion of values that lie below this z-score. The table gives us the proportion of the distribution that is to the left of the z-score. A z-score of -1.5 corresponds to a proportion of approximately 0.0668.
Therefore, the proportion of soda cans containing less than 12 ounces is approximately 0.0668, or 6.68%.
How do I write the equation of a line in slope intercept form, where slope is -3 and the y-intercept is (0,-10)
Answer:
y= -3x +10
Step-by-step explanation:
you already have your slop so plug it in and your y intercept is 10