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2 years ago

7

William has three $10 bills and four $1 bills in his wallet. What is the probability that he will reach into his wallet twice an

d pull out a $10 bill each time? Assume he does not replace the first bill when he pulls out the second one.

1 answer:

bulgar [2K]2 years ago

7 0

Answer:

Step-by-step explanation can’t because he might pull out more than 10 dollars or less than 10 dollars.

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In the past, 19% of all homes with a stay-at-home parent had the father as the stay-at-home parent. An independent research firm

Oksi-84 [34.3K]

Answer:

A sample size of 657 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

In the past, 19% of all homes with a stay-at-home parent had the father as the stay-at-home parent.

This means that $\pi = 0.19$

(a) What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03?

A sample size of n is needed.

n is found when $M = 0.03$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.19*0.81}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.19*0.81}$

$\sqrt{n} = \frac{1.96\sqrt{0.19*0.81}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.19*0.81}}{0.03})^{2}$

$n = 656.91$

Rounding up to the nearest whole number.

A sample size of 657 is needed.

5 0

3 years ago

What is the slope of the line given by the equation y = -5x? Enter your answer as an integer or fraction in lowest terms.

posledela

Y=mx+b
m= slope
so -5 is the slope of the given line

5 0

3 years ago

A researcher collects data on the relationship between the amount of daily exercise an individual gets and the percent body fat

fredd [130]

Answer: approximately 24

Step-by-step explanation:

We need to plot a regression line.

So we fit a model using the regression of Y on X, that an equation that predict Y for a given X using:

(Y -mean(Y ))= a(X-meanX)...........1

Where the formular of a is given the attachment.

N= the of individuals = 5

Y = amount of fat

X = time of exercise

mean(X )= sum of all X /N

= 131/5 = 26.2

mean(Y) = sum of all Y/N

= 104/5 = 20.8

a = N(SXY) - (SX)(SY)/ NS(X²) -(SX)²......2

SXY = Sum of Product X and Y

SX= sum of all X

SY = Sum of all Y

S(X²)= sum of all X²

(SX) = square of sum of X

a = -0.478

Hence we substitute into 1

Y-20.8 = -0.478 (X-26.2)

Y -20.8 = -0.478X - 12.524

Y = -0.478X + 33.324 or

Y = 33.324 - 0.478X (model)

When X = 20

Y = 33.324 - 0.478 × 20

Y = 33.324 - 9.56

Y = 23. 764

Y =24(approximately)

Carefully meaning of formula used in attachment to the solution they are the same.

4 0

3 years ago

Given a mean of 150 and a standard deviation of 25 Define the interval about the mean that contains 95% of the data.

ohaa [14]

By normal curve symmetry

<span>from normal table </span>
<span>we have z = 1.15 , z = -1.15 </span>

<span>z = (x - mean) / sigma </span>
<span>1.15 = (x - 150) / 25 </span>
<span>x = 178.75 </span>

<span>z = (x - mean) / sigma </span>
<span>-1.15 = (x - 150) / 25 </span>
<span>x = 121.25 </span>

<span>interval is (121.25 , 178.75) </span>

<span>Pr((121.25-150)/25 < x < (178.75-150)/25) </span>
<span>is about 75%</span>

7 0

2 years ago

Use the decimals 2.47, 9.57, and 7.1 to write two different addition facts and two different subtraction facts

kaheart [24]

2.47+7.1=9.57
7.1+2.47=9.57
9.57-2.47=7.1
9.57-7.1=2.47

3 0

3 years ago