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
The price of a new computer is p dollars. The computer is on sale for 30% off. Which expression shows the savings that are being offered on the computer?
A. p - 0.3p B. 0.7 × p C. 0.3 × p D. p ÷ 30
Option C
Expression that shows the savings that are being offered on the computer is 0.3p
Solution:
Given that price of a new computer is p dollars
The computer is on sale for 30% offer
To find: Expression that shows the savings that are being offered on the computer
Computer is on sale for 30% offer which means 30 % offer on original price "p"
Original price = "p" dollars
offer price / saved price = 30 % of "p"
[tex]\text{ saved price } = 30 \% \times p\\\\\text{ saved price } = \frac{30}{100} \times p\\\\\text{ saved price } = 0.3p[/tex]
Thus the required expression is 0.3p
Thus option C is correct.
A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. The following morning she finds four pennies. Apparently, while she sleeps each penny turns into two magic pennies. The total number of pennies seen under the pillow each day is the grand total; that is, the pennies from each of the previous days are not being stored away until more pennies magically appear. How many days would elapse before she has a total of more than $2 billion ? (Proceed by trial and error.)
nothing days
(Type a whole number.) PLEASE HURRY
Answer: 31
Step-by-step explanation:
2^x=2 000 000 000
log2^x=log2 000 000 000
xlog2 = log 2 000 000 000
x= log (2000 000 000)/log 2
x= 30.897352854
round to 31
gotchu bro
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
11.8 7.7 6.5 6.8 9.7 6.8 7.3
7.9 9.7 8.7 8.1 8.5 6.3 7.0
7.3 7.4 5.3 9.0 8.1 11.3 6.3
7.2 7.7 7.8 11.6 10.7 7.0
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. [Hint: ?xi = 219.5.] (Round your answer to three decimal places.)
MPa
State which estimator you used.
x
p?
s / x
s
x tilde
Answer:
The point estimate for population mean is 8.129 Mpa.
Step-by-step explanation:
We are given the following in the question:
Data on flexural strength(MPa) for concrete beams of a certain type:
11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0
a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured
We use the sample mean, [tex]\bar{x}[/tex] as the point estimate for population mean.
Formula:
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129[/tex]
Thus, the point estimate for population mean is 8.129 Mpa.
To estimate the mean flexural strength, the sum of strengths (219.5 MPa) is divided by the total number of beams measured (26), which yields a mean value of 8.442 MPa when rounded to three decimal places. The estimator used is the sample mean.
Explanation:To calculate a point estimate of the mean value for flexural strength (MPa) for a conceptual population of concrete beams, we use the sum of all measured strengths and divide by the number of measurements. The sum of the flexural strengths is provided as Σxi = 219.5 MPa.
Given the dataset:
11.87.76.56.89.76.87.37.99.78.78.18.56.37.07.37.45.39.08.111.36.37.27.77.811.610.77.0The number of measurements is the number of data points, which is 26. To find the mean:
mean = Sum of strengths / Number of measurements
mean = 219.5 MPa / 26
mean = 8.442 MPa (rounded to three decimal places)
The estimator used here is the sample mean (×).
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Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. Loan officers assess information about the applicant, totalling points they award for the persons income level, credit history, current debt burden, and so on. The higher the point total, the more convinced the bank is that it’s safe to make the loan. Any applicant with a lower point total than a certain cut-off score is denied a loan. We can think of this decision as a hypothesis test. Since the bank makes its profit from the interest collected on repaid loans, their null hypothesis is that the applicant will repay the loan and therefore should get the money. Only if the persons score falls below the minimum cut-off will the bank reject the null and deny the loan. This system is reasonably reliable, but, of course, sometimes there are mistakes.a) When a person defaults on a loan, which type of error did the bank make?b) Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it?c) Suppose the bank decides to lower the cut-off score from 250 points to 200. Is that analogous to choosing a higher or lower value of for a hypothesis test? Explain.d) What impact does this change in the cut-off value have on the chance of each type of error?
Answer:
(a) Type II error
(b) Type I error
(c) It is analogous to choosing a lower value for a hypothesis test
(d) There will be more tendency of making type II error and less tendency of making type I error
Step-by-step explanation:
(a) The bank made a type II error because they accepted the null hypothesis when it is false
(b) The bank made a type I error because they rejected the null hypothesis when it is true
(c) By lowering the value for the hypothesis test, they give applicants who do not meet the initial cut-off point the benefit of doubt of repaying the loan thus increasing their chances of making more profit
(d) There will be more tendency of making type II error because the bank accepts the null hypothesis though they are not fully convinced the applicants will repay the loan and less tendency of making type I error because the bank rejects the null hypothesis knowing the applicants might not be able to repay the loan
In hypothesis testing, a person defaulting on a loan represents a Type I error, while missing an opportunity to make a loan to someone who would have repaid it represents a Type II error. Lowering the cut-off score is analogous to increasing the value in a hypothesis test, accepting more risk. This increases the likelihood of Type I errors but decreases the likelihood of Type II errors.
Explanation:In the context of hypothesis testing in banking and the financial capital market, (a) when a person defaults on a loan, the bank made a Type I error: they lent money to an individual who failed to repay it. (b) If the bank does not lend money to someone who would have repaid it, it's a Type II error: they missed an opportunity to profit from interest because they incorrectly predicted the person would not pay back the loan. (c) Lowering the cut-off score from 250 points to 200 is analogous to choosing a higher value for a hypothesis test, which means the bank is willing to accept more risk. (d) Changing the cut-off value impacts the chance of each kind of error. By lowering the score, the bank is more likely to make Type I errors (lending to individuals who won't repay), but less likely to make Type II errors (not lending to individuals who would repay).
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A researcher developing scanners to search for hidden weapons at airports has concluded that a new scanner is significantly better than the current scanner. He made his decision based on a test using alpha equals 0.025 . Would he have made the same decision at alpha equals 0.10 question mark How about alpha equals 0.005 question mark Explain.
Step-by-step explanation:
Since the decision is made on the test based on the use of alpha equals 0.025, the p-value of the test would have been higher than the level of significance provided that is 0.025 since the test is not important.
p > 0.025
Now if we know that p > 0.025, this would not necessarily mean that p > 0.1 also, therefore we do not know with the given information that he would have made the same decision for 0.1 level of significance, ( we are not sure about his decision in that case ).
Now for the level of significance of 0.005, we would be sure that p > 0.005 as it is greater than 0.025, therefore the test is not significant at this level of significance as well. Therefore he would have made the same decision for 0.005 level of significance.
For the Data Set below, calculate the Variance to the nearest hundredth decimal place. (Do not use a coma in your answer) 175 349 234 512 638 549 500 611
Answer:
The variance of the data is 29966.3.
Step-by-step explanation:
The given data set is
175, 349, 234, 512, 638, 549, 500, 611
We need to find the variance to the nearest hundredth decimal place.
Mean of the data
[tex]Mean=\dfrac{\sum x}{n}[/tex]
where, n is number of observation.
[tex]Mean=\dfrac{3568}{8}=446[/tex]
The mean of the data is 446.
[tex]Variance=\dfrac{\sum (x-mean)^2}{n-1}[/tex]
[tex]Variance=\dfrac{(175-446)^2+(349-446)^2+(234-446)^2+(512-446)^2+(638-446)^2+(549-446)^2+(500-446)^2+(611-446)^2}{8-1}[/tex]
[tex]Variance=\dfrac{209764}{7}[/tex]
[tex]Variance=29966.2857[/tex]
[tex]Variance\approx 29966.3[/tex]
Therefore, the variance of the data is 29966.3.
Final answer:
The variance of the given data set is calculated by finding the mean, squaring the differences from the mean, summing these squares, and dividing by the count minus one. It results in a variance of 12790.54 when rounded to the nearest hundredth decimal place.
Explanation:
To calculate the variance of the data set, follow these steps:
First, find the mean (average) of the data set by adding all the numbers together and dividing by the total count.
Next, subtract the mean from each data point and square the result to get the squared differences.
Then, add up all of the squared differences.
Finally, divide the sum of the squared differences by the total number of data points minus one to get the variance (since this is a sample variance).
Data Set: 175, 349, 234, 512, 638, 549, 500, 611
Mean = (175 + 349 + 234 + 512 + 638 + 549 + 500 + 611) / 8 = 3793 / 8 = 474.125
Squared differences = (175 - 474.125)^2 + (349 - 474.125)^2 + (234 - 474.125)^2 + (512 - 474.125)^2 + (638 - 474.125)^2 + (549 - 474.125)^2 + (500 - 474.125)^2 + (611 - 474.125)^2
Sum of squared differences = 89533.78125
Variance = 89533.78125 / (8 - 1) = 12790.54
Therefore, the variance of the data set, to the nearest hundredth decimal place, is 12790.54.
If n is a positive integer, which of following statement is individually sufficient to prove whether 289 is a factor of n?a. The greatest common divisor of n and 344 is 86. b. Least common multiple of n and 272 is 4624. c. The least common multiple of n and 289 is 289n.
Answer:
The statement b) is individually sufficient to prove than 289 is a factor of n
Step-by-step explanation:
The least common multiple of n and 272 is the smallest number that is a multiple of n and a multiple of 272. Therefore:
272 x X = 4624 ⇒ X = 17 but 272 = 17 · 16 and 289 = 17 · 17
Therefore 17·17 must be a factor of n. That means 289 is a factor of n
Suppose that n(U) = 200, n(A) = 105, n(B) = 110, and n( A ∩ B ) = 30. Find n( A c ∪ B ). a) 80 b) 110 c) 125 d) 95 e) 75 f) None of the above.
Answer:
Option c - [tex]n(A^c\cup B)=125[/tex]
Step-by-step explanation:
Given : Suppose that n(U) = 200, n(A) = 105, n(B) = 110, and n(A∩B) = 30.
To find : The value of [tex]n(A^c\cup B)[/tex] ?
Solution :
n(U) = 200, n(A) = 105, n(B) = 110, and n(A∩B) = 30
We know that,
[tex]n(A^c)=n(U)-n(A)[/tex]
[tex]n(A^c)=200-105[/tex]
[tex]n(A^c)=95[/tex]
and [tex]n(A^c \cap B)=n(B)-n(A\cap B)[/tex]
[tex]n(A^c \cap B)=110-30[/tex]
[tex]n(A^c \cap B)=80[/tex]
Now, [tex]n(A^c\cup B)=n(A^c)+n(B)-n(A^c \cap B)[/tex]
[tex]n(A^c\cup B)=95+110-80[/tex]
[tex]n(A^c\cup B)=125[/tex]
Therefore, option c is correct.
The value of the union set given as n(A^c U B) is; C: 125
What is the union of the set?
We are given;
n(U) = 200, n(A) = 105, n(B) = 110, and n(A ∩ B) = 30.
In sets, we know that complement of set A is;
n(A^c) = n(U) - n(A)
Thus; n(A^c) = 200 - 105
n(A^c) = 95
Also, we know that;
n(A^c ∩ B) = n(B) - n(A ∩ B)
n(A^c ∩ B) = 110 - 30
n(A^c ∩ B) = 80
Thus;
n(A^c U B) = n(A^c) + n(B) - n(A^c ∩ B)
n(A^c U B) = 95 + 110 - 80
n(A^c U B) = 125
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Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year. What is the dividend yield?
A. 3.75 percent
B. 4.20 percent
C. 4.55 percent
D. 5.25 percent
E. 6.60 percent
Answer:
E. 6.60 percent
Step-by-step explanation:
We have been given that Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year.
We will use total return formula to answer our given problem.
[tex]\text{Total return}=\text{Dividend yield}+\text{Growth rate}[/tex]
Upon substituting our given values in above formula, we will get:
[tex]9.2\%=\text{Dividend yield}+2.6\%[/tex]
[tex]\text{Dividend yield}=9.2\%-2.6\%[/tex]
[tex]\text{Dividend yield}=6.6\%[/tex]
Therefore, the dividend yield would be 6.60% and option E is the correct choice.
The lumen output was determined for each of I = 3 different brands of lightbulbs having the same wattage, with J = 8 bulbs of each brand tested. The sums of squares were computed as SSE = 4776.3 and SSTr = 599.5. State the hypotheses of interest (including word definitions of parameters).
Answer:
Step-by-step explanation:
Hello!
The study variable is
X: Lumen of a bulb of the i brand. i=3
There are 3 populations of bulbs, Brand 1, Brand 2 and brand 3.
The objective is to test if the population means are equal.
The study parameters are:
μ₁: population mean lumen of the population of light bulbs of brand 1.
μ₂: population mean lumen of the population of light bulbs of brand 2.
μ₃: population mean lumen of the population of light bulbs of brand 3.
The hypothesis is:
H₀:μ₁= μ₂= μ₃= μ
H₁: At least one of the population means is different.
To test this hypothesis, considering the given information, I'll use an ANOVA test, then the statistic is defined as:
[tex]F= \frac{MSTr}{MSerror}[/tex]~[tex]F_{(I-1)(J-1)}[/tex]
Rejection region
This region is always one-tailed (right), the statistic is constructed as the mean square of the treatments divided by the mean square of the error, if the number of F is big, this means that the treatments have more effect over the populations. If the value of F is small, this means that there is no difference between the variability caused by the treatments and the one caused by the residues.
Since there is no significance level specified, I'll use α: 0.05
[tex]F_{(I-1);(J-1); 1 - \alpha } = F_{2; 7; 0.95} = 19.35[/tex]
You will reject the null hypothesis when F[tex]_{H_0}[/tex] ≥ 19.35
To calculate the statistic value you need to calculate the Mean Square of Treatments and the Mean Square of errors:
MSTr= SSTr/DfTr = 599.5/2= 299.75
MSerror= SSerror/Dferror= 4776.3/5= 955.26
F[tex]_{H_0}[/tex]= [tex]\frac{299.75}{955.26}[/tex]= 0.31
At this level the decision is to not reject the null hypothesis.
I hope it helps!
Consider a manufacturing process with a quality inspection station. In the past, 15% of parts are defective. As soon as one defective part is found, the process is stopped. If 8 parts have been inspected without finding a defective part, what is the probability that at least 11 total parts will be inspected before the process is stopped?
Answer:
0.614125
Step-by-step explanation:
Given that a manufacturing process with a quality inspection station has on an average 15% of parts are defective.
As soon as one defective part is found, the process is stopped.
We find that number of defectives would be binomial because each part randomly selected has a constant probability of 0.15 being defective
Probability that at least 11 total parts will be inspected before the process is stopped/8 parts have been inspected without finding a defective part
=[tex]P(x\geq 11)/P(x=8)\\[/tex]
= Probability of 9th, 10th, 11th should not be defective
= [tex](1-0.15)^3\\= 0.614125[/tex]
The concept of aging as a result of cellular duplication errors is based on the fact that the body's ability to make new cells that are exact copies of the original ones: a. stops at about age 80. b. continues unchanged throughout life. c. becomes less able to repair duplication errors. d. stops at about age 18.
Answer:
Becomes less able to repair duplication errors
Step-by-step explanation:
This is premised on the fact that aging has been connected with the deterioration of DNA maintenance and repair machinery, which tends to lose its ability to replicate new cell as a person age with time.
If SSXY = −16.32 and SSX = 40.00 for a set of data points, then what is the value of the slope for the best-fitting linear equation? a. −0.41 b. −2.45 c. positive d. There is not enough information; you would also need to know the value of SSY.
Answer: a. −0.41
Step-by-step explanation:
The slope for the best-fitting linear equation is given by :-
[tex]b=\dfrac{SS_{xy}}{SS_x}[/tex]
where , [tex]SS_x[/tex] =sum of squared deviations from the mean of X.
[tex]SS_{xy}[/tex] = correlation between y and x in terms of the corrected sum of products.
As per given , we have
[tex]SS_x=10.00[/tex]
[tex]SS_{xy}=-16.32[/tex]
Then, the value of the slope for the best-fitting linear equation will be
[tex]b=\dfrac{-16.32}{40.00}=-0.408\approx -0.41[/tex]
Hence, the value of the slope for the best-fitting linear equation= -0.41
So the correct answer is a. −0.41 .
The value of the slope for the best-fitting linear equation is -0.41
The given parameters are:
[tex]SS_{xy} = -16.32[/tex] --- the correlation between y and x
[tex]SS_{x} = 40.00[/tex] --- the sum of squared deviations from the mean of X.
The slope (b) is calculated using the following formula
[tex]b = \frac{SS_{xy}}{SS_x}[/tex]
Substitute values for SSxy and SSx
[tex]b = \frac{-16.32}{40.00}[/tex]
Divide -16.32 by 40.00
[tex]b = -0.408[/tex]
Approximate
[tex]b = -0.41[/tex]
Hence, the value of the slope for the best-fitting linear equation is -0.41
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A bag contains 8 red marbles, 3 blue marbles and 6 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 1000th, that all three marbles drawn will be red?
Answer:
0.082
Step-by-step explanation:
There are a total of 17 marbles, 8 of which are red.
The probability that the first marble is red is 8/17.
The probability that the second marble is red is 7/16.
The probability that the third marble is red is 6/15.
Therefore, the probability that all three marbles are red is:
P = 8/17 × 7/16 × 6/15
P = 7/85
P = 0.082
You perform a X2 goodness-of-fit test to see if the number of birthdays occurring each month matches the expected number (assuming each month is equally likely to be the birth month for any given individual). You get 20.5 as your X2 value. What is the P-value for this test?
Answer:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: Each month is equally likely to be the birth month for any given individual
H1: Each month is NOT equally likely to be the birth month for any given individual
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
After calculate the statistic we got [tex]\chi^2 = 20.5[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=categories-1=12-1=11[/tex]
And we have categories =12 since we have 12 months in a year
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
Identify the type of observational study (cross-sectional, retrospective, or prospective) described below. A research company uses a device to record the viewing habits of about 2500 households, and the data collected over the past 2 years will be used to determine whether the proportion of households tuned to a particular children's program increased. Which type of observational study is described in the problem statement?
A. A prospective study
B. A retrospective study
C. A cross-sectional study
Answer:
B
Step-by-step explanation:
The retrospective or historic cohort story, is a longitudinal cohort story that considers a particular set of individuals that share the same exposure factor to ascertain its influence in the developments of an occurrence, which are compared with the other set or cohort which were not exposed to the same factors.
Retrospective studies have existed about the same time as prospective studies, hence their names.
A psychologist wants to see if a certain company has fair hiring practices in an industry where 60% of the workers are men and 40% are women. She finds that the company has 55 women and 52 men. Test to see if these numbers are different from the industry numbers, and if so, how are they different? Use alpha -.05 and four steps. A) what is the null hypothesis? B) what is the alternative hypothesis? C) what is the critical value of the test statistic? D) what is the value of the test statistic? E) Reject or accept the null? And why?
The hypothesis test examines if the company's hiring distribution differs from industry standards. The null hypothesis represents no difference, while the alternative suggests a discrepancy.
The critical value for the test statistic at a 0.05 significance level is ±1.96 for a two-tailed test, and we either reject or fail to reject the null based on the comparison of the calculated Chi-square statistic to the critical value.
To determine if there is a significant difference between the hiring practices of a certain company and the industry standard, we use a hypothesis test for proportions.
A. Null Hypothesis (H₀)
The null hypothesis H0: P_(men) = 0.60 and P_(women) = 0.40, where P represents the proportion of men and women in the company, respectively.
B. Alternative Hypothesis (Ha)
The alternative hypothesis Ha: P_(men) ≠ 0.60 and P_(women) ≠ 0.40.
C. Critical Value of Test Statistic
The critical value for a two-tailed test at alpha = 0.05 is z = ±1.96.
D. Value of the Test Statistic
To calculate the test statistic, we use the formula for a test of proportions:
Calculate the expected counts based on industry proportions: expected men = 107 * 0.60 = 64.2, expected women = 107 * 0.40 = 42.8.
Compute the Chi-square test statistic: Χ2 = ((52-64.2)2/64.2) + ((55-42.8)2/42.8).
The resulting Χ₂ statistic can then be compared against the critical Χ₂ value with 1 degree of freedom at alpha = 0.05, which is 3.841.
E. Reject or Accept the Null Hypothesis
If the calculated Χ₂ is greater than 3.841, we reject the null hypothesis; if not, we fail to reject the null hypothesis. Without the actual calculation of the Χ₂, we cannot definitively conclude the action on the null hypothesis in this context.
The random variable X = the number of vehicles owned. Find the expected number of vehicles owned. Round answer to two decimal places.
Answer:
The expected number of vehicles owned to two decimal places is: 1.85.
Step-by-step explanation:
The table to the question is attached.
[tex]E(X) =[/tex]∑[tex]xp(x)[/tex]
Where:
E(X) = expected number of vehicles owned
∑ = Summation
x = number of vehicle owned
p(x) = probability of the vehicle owned
[tex]E(X) = (0 * 0.1) + (1 * 0.35) + (2 * 0.25) + (3 * 0.2) + (4 * 0.1)\\E(X) = 0 + 0.35 + 0.50 + 0.60 + 0.4\\E(X) = 1.85[/tex]
The expected number of vehicles owned is 1.85.
The expected number of vehicles owned, based on probability of ownership of 0 to 3 vehicles, is calculated by multiplying each possible number of vehicles by their corresponding probabilities and then summing up all the products. The calculated expected number is approximately 1.7 vehicles.
Explanation:To find the expected number of vehicles owned, we first need to multiply each possible number of vehicles someone could own by the probability of them owning that many vehicles. Then, sum up all of these products.
For instance, if they could own up to 3 cars and the probability for owning 0, 1, 2, or 3 cars is 0.1, 0.3, 0.4, and 0.2 respectively:
For 0 cars: 0 * 0.1 = 0
For 1 car: 1 * 0.3 = 0.3
For 2 cars: 2 * 0.4 = 0.8
For 3 cars: 3 * 0.2 = 0.6
Adding these together gives the expected number of cars:
0 + 0.3 + 0.8 + 0.6 = 1.7 (rounded to two decimal places).
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2. I Using the example { 2/3+4/3 X, explain why we add fractions the way we do. What is the logic behind the procedure? Make math drawings to support your explanation
Answer:
The procedure emphasizes the idea of the summation of one physical quantity. In this case, X.
Step-by-step explanation:
1. When we add fractions like these we do it simply by rewriting a new one, the summation of the numerators over the same denominator:
[tex]\frac{2}{3}X+\frac{4}{3})X=\frac{6}{3}X= 2X[/tex]
The procedure emphasizes the idea of the summation of one physical quantity, in this case, X.
2) This physical quantity x could be miles, oranges, gallons, etc.
Tyler has a baseball bat that weighs 28 ounces. Find this weight in kilograms and in grams. (Note 1 kilogram=35 ounces)
Answer:0.8 kilograms
800 grams
Step-by-step explanation:
The weight of Tyler's baseball bat is 28 ounces. We would convert the weight in ounces to kilogram and grams.
Let x represent the number of kilograms that is equal to 28 ounces. Therefore
1 kilogram = 35 ounces
x kilogram = 28 ounces
Cross multiplying, it becomes
35 × x = 28 × 1
35x = 28
x = 28/35 = 0.8 kilograms
We would convert 0.8 kilograms to grams
Let y represent the number of grams that is equal to 0.8 kilograms. Therefore,
1000 grams = 1 kilogram
y grams = 0.8 kilograms
Cross multiplying,
y × 1 = 0.8 × 1000
y = 800 grams
Answer:
0.2
Step-by-step explanation:
A random sample of 100 high school students was surveyed regarding their favorite subject. The following counts were obtained: Favorite Subject Number of Students English Math Science 30 Art/Music The researcher conducted a test to determine whether the proportion of students was equal for all four subjects. What is the value of the test statistic? O b. 25 OOOO d. -4 How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories? a. 74 OOOO d. 62
Answer:
a) [tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
b) [tex]df=Categories-1=10-1=9[/tex]
Step-by-step explanation:
We assume the following info:
Favorite Subject Number of students
English 25
Math 30
Science 30
Art/Music 15
Total 100
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Part a
The system of hypothesis on this case are:
H0: There is no difference with the distribution proposed
H1: There is a difference with the distribution proposed
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values are 25 for each category.
And the calculations are given by:
[tex]E_{English} =25[/tex]
[tex]E_{Math} =25[/tex]
[tex]E_{Science} =25[/tex]
[tex]E_{Music} =25[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=Categories-1=4-1=3[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >6)=0.112[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(6,3,TRUE)"
Part b
For this case we have this formula:
[tex]df=Categories-1=10-1=9[/tex]
Suppose that ten bats were used in the experiment. For each trail, the zoo keeper pointed to one of two "feeders". Suppose that the bats went to the correct feeder (the one that the zoo keeper pointed at) 8 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point. (0.62, 1.0) (0.477, 0.951) (0.321, 0.831)
Answer: (0.477, 0.951)
Step-by-step explanation:
Given : Number of observations : n = 10
Number of successes : x = 8
Let p be the population proportion of times that the bats would follow the point.
Because the number of observation is not enough large , so we use plus four confidence interval for p.
Plus four estimate of p=[tex]\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}[/tex]
[tex]\hat{p}=\dfrac{8+2}{10+4}=\dfrac{10}{14}\approx0.714[/tex]
We know that , the critical value for 95% confidence level : z* = 1.96 [By using z-table]
Now, the required confidence interval will be :
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 14
[tex]0.714\pm (1.96)\sqrt{\dfrac{0.714(1-0.714)}{14}}[/tex]
[tex]0.714\pm (1.96)\sqrt{0.014586}[/tex]
[tex]0.714\pm (1.96)(0.120772513429)[/tex]
[tex]\approx0.714\pm0.237=(0.714-0.237,\ 0.714+0.237)[/tex]
[tex](0.477,\ 0.951)[/tex]
Hence, the 95% confidence interval for the population proportion of times that the bats would follow the point = (0.477, 0.951)
The 95% confidence interval for the proportion of the times that bats would follow the point is (0.552, 1.0). The result was adjusted because proportions cannot exceed 1.
Explanation:To calculate the 95% confidence interval for the population proportion, we follow these steps:
First, we calculate the sample proportion (p-hat) as the number of successes (bats going to the correct feeder) divided by the total number of observations. In this case, p-hat = 8 ÷ 10 = 0.8.Next, our goal is to construct the confidence interval using the formula p-hat ± Z * sqrt[p-hat(1 - p-hat) / n], where Z is the Z-value in the standard normal distribution corresponding to the desired confidence level (1.96 for 95% confidence level), n is the number of observations, and p-hat is the calculated sample proportion.Substituting all values into the formula, we get 0.8 ± 1.96 * sqrt[0.8(0.2) / 10] = 0.8 ± 1.96 * 0.126 = (0.552, 1.048).However, this interval contains value bigger than 1, which is not possible because proportion cannot exceed 1.Hence, we adjust our interval to (0.552, 1.0).Learn more about Confidence Interval here:https://brainly.com/question/34700241
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The mean time taken to design a house plan by 38 architects was found to be 22 hours with a standard deviation of 3.70 hours. a. Construct a 95% confidence interval for the population mean μ. Round your answers to two decimal places.
To construct a 95% confidence interval for the population mean of the time taken to design a house plan, use the formula which states 21.08 to 22.92 hours.
Explanation:To construct a 95% confidence interval for the population mean, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation/sqrt(sample size))
Given that the mean time taken to design a house plan is 22 hours, the standard deviation is 3.70 hours, and the sample size is 38, we can calculate the confidence interval:
Confidence Interval = 22 ± (1.96) * (3.70/sqrt(38))
Calculating this gives us a confidence interval of approximately 21.08 to 22.92 hours.
Learn more about Confidence Intervals here:https://brainly.com/question/34700241
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A supervisor records the repair cost for 11 randomly selected refrigerators. A sample mean of $82.43 and standard deviation of $13.96 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Final answer:
The critical value for constructing a 99% confidence interval is 2.576.
Explanation:
To determine the critical value for constructing the 99% confidence interval, we need to find the Z-value that represents the level of confidence. For a 99% confidence interval, the alpha level (1 - confidence level) is 0.01. Since the data is approximately normally distributed and the sample size is greater than 30, we can use the Z-distribution. Using a Z-table or calculator, we find that the Z-value for a 0.01 alpha level is approximately 2.576.
Which of the following is used to determine the significance of predictions made by a best fitting linear equation?A. correlational analysisB. analysis of varianceC. analysis of regressionD. method of least squares
Answer:
D. method of least squares
Step-by-step explanation:
The Least Squares Method (LSM) is a mathematical method used to solve various problems, based on minimizing the sum of the squared deviations of some functions from the desired variables. It can be used to “solve” over-determined systems of equations (when the number of equations exceeds the number of unknowns), to find a solution in the case of ordinary (not redefined) linear or nonlinear systems of equations, to approximate the point values of a function. OLS is one of the basic regression analysis methods for estimating the unknown parameters of regression models from sample data.
Correlation analysis is a statistical method used to assess the strength of the relationship between two quantitative variables. A high correlation means that two or more variables have a strong relationship with each other, while a weak correlation means that the variables are hardly related. In other words, it is a process of studying the strength of this relationship with available statistics.
Analysis of Variance (or ANOVA) is a collection of statistical models used to analyze group averages and related processes (such as intra- and inter-group variation) in statistical science. When using Variance Analysis, the observed variance of a specified variable is divided into the variance component that can be based on different sources of change. In its simplest form, "Analysis of Variance" is a inferential statistical test to test whether the averages of several groups are equal or not, and this test generalizes the t-test test for two-groups to multiple-groups. If multiple two-sample-t-tests are desired for multivariate analysis, it is clear that this results in increased probability of type I error. Therefore, the variance analysis would be more useful to compare the statistical significance of three or more means (for groups or for variables) with the test.
Regression analysis is an analysis method used to measure the relationship between two or more variables. If analysis is performed using a single variable, it is called univariate regression, and if more than one variable is used, it is called multivariate regression analysis. With the regression analysis, the existence of the relationship between the variables, if there is a relationship between the strength of the information can be obtained. The logic here is that the variable to the left of the equation is affected by the variables to the right. The variables on the right are not affected by other variables. Not being influenced here means that when we put these variables into a linear equation in mathematical sense, it has an effect. Multiple linearity, sequential dependency problems are not meant.
A professor is trying to determine if her students guessed on a certain multiple choice question. She expects that if the students guessed, the distribution of answers would be uniform for that question. She compares the observed distribution of answers with the uniform distribution. The professor conducts a chi-square Goodness-of-Fit hypothesis test at the 5% significance level.
a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution.
b) χ20=13.167.
c) χ20.05=7.815.
d) What conclusions can be made?
Select all that apply:
A. We should reject H0.
B. We should not reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
D. At the 5% significance level, there is not enough evidence to conclude that the students were not guessing.
Answer:
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: The student answers have the uniform distribution.
H1: The student answers do not have the uniform distribution.
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
On this case we assume that the calculated statistic is given by:
Statistic calculated
[tex]\chi^2_{calc}=13.167[/tex]
P value
Assuming the we have 2 rows and 4 columns on the contingency table.
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(2-1)(4-1)=3[/tex]
We can calculate the critical value with this formula in excel:" =CHISQ.INV(0.95,3)" On this case we got that the critical value is:
[tex]\chi^2_{crit}=7.815[/tex]
Since our calculated value is higher than the cirtical value we have enough evidence to reject the null hypothesis at the significance level of 5%.
And we can also calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >13.167)=0.0043[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(13.167,3,TRUE)"
Since the p value is lower than the significance level we reject the null hypothesis at 5% of significance.
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
The reason why we select option C is because if we reject the null hypothesis of uniform distribution then we are rejecting the claim that the students are guessing.
A box contains the following numbered tickets: 1,1,5,9,9
a) If I draw two tickets with replacement, what is the chance that the sum of the two tickets is greater than or equal to 10?
b) Drawing three tickets without replacement, what is the chance the first two tickets are not 5's, and the last ticket is a 5?
c) Calculate b) if the draws are made with replacement.
d) If I repeat the procedure in a) 8 times (ie draw 2 tickets and find their sum, and do this 8 times), what is the chance that I get a sum greater than or equal to 10 exactly 6 of the 8 times?
Answer:
Step-by-step explanation:
Feel free to ask if anything is unclear
An advertisement for a word-processing class claims that students who complete the class will, on average, be able to type 60 words per minute (wpm) with a standard deviation of 6 wpm. At the end of the class, 49 students are tested and their mean score is 58.5 wpm.
A. Is this evidence that the true mean is different from 60 wpm? Give a complete answer, using a significance level of .05. (12 points)
B. Suppose the person conducting the research had believed, before collecting data, that the graduates weren't as good as claimed. Would this belief have changed the analysis you performed in part (A)? (12 points)
C. Construct a 99% confidence interval for the true mean of the population. What conclusion can you draw from this interval, and do you have evidence to reject the claim that the average graduate can type 60 wpm? (6 points)
Answer:
Reject at 5%, accept at 1% the null hypothesis
Step-by-step explanation:
Set up hypotheses as
[tex]H_0: \bar x = 60\\H_a: \bar x < 60[/tex]
(Left tailed test)
Population std dev = 6
Sample std error = [tex]\frac{6}{\sqrt{49} } \\=0.8555[/tex]
Mean difference = -1.5
Since sigma is known we can use Z test
Z = mean diff/std error = -1.7533
p value = 0.039
a) Since p value <0.05 we reject H0. There is evidence that the true mean is different from 60 wpm
b) Yes, because his sample would have been biased since he may want to prove his belief so slow or inefficient persons he would have selected in the sample.
c) For 99% confidence interval critical value = 2.58
Confidence interval for population mean = 58.5±2.58*std error
=(56.2928, 60.7072)
Since this contains 60, the hypothesized mean, we accept null hypothesis.
we do not have evidence to reject the claim that the average graduate can type 60 wpm at 1% level of significance.
An article reported that for a sample of 58 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.4.
(a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.) , ppm Interpret the resulting interval. We are 95% confident that the true population mean lies below this interval. We are 95% confident that this interval does not contain the true population mean. We are 95% confident that this interval contains the true population mean. We are 95% confident that the true population mean lies above this interval.
(b) Suppose the investigators had made a rough guess of 184 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 47 ppm for a confidence level of 95%?
Answer:
Step-by-step explanation:
Juan roller skates at the constant speed of 8 miles per hour. How far can he travel in 2.4 hours.
Answer: Juan can travel 19 miles in 2.4 hours at a speed of 8 miles per hour
Step-by-step explanation:
Juan roller skates at the constant speed of 8 miles per hour. Distance travelled is expressed as
Distance = speed × time
Therefore, the distance that Juan can travel in 2.4 hours is
Distance = 2.4 × 8 = 19.2 miles
Approximating to the nearest whole number, it becomes 19 miles