Refer to the triangle. Round to the nearest The value of x is tenth.

Robert sketches two rectangular prisms, A and B. Prism A's side lengths are 5 centimeters, 6 centimeters, and 7 centimeters. Pri

erik [133]

Answer:

90

Step-by-step explanation:

10+67

7586+65

576845=8757685

2+88=90

4 0

3 years ago

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confid

katrin [286]

Answer:

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 865, \pi = \frac{408}{865} = 0.4717$

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 $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 - 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.4384$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 + 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.5050$

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

6 0

3 years ago

The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per w

alukav5142 [94]

Answer:

The standard deviation of number of hours worked per week for these workers is 3.91.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem we have that:

The average number of hours worked per week is 43.4, so $\mu = 43.4$ .

Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers.

This means that the Z score of $X = 48$ has a pvalue of 0.88. This is Z between 1.17 and 1.18. So we use $Z = 1.175$ .