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3 years ago

Hazel covered a box with felt. The dimensions of the box are shown in the figure below:

1 answer:

8 0

<span>$36.36 because the area of the prism is 404sq. in. , 404x0.09 (for nine cents per sq. in.) = 36.36
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Hope this helps!! :)

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Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

2 years ago

A researcher surveyed 150 high school students and found that 68% played a musical insturment. What would be a reasonable range

maria [59]

Answer:

The reasonable range for the population mean is (61%, 75%).

Step-by-step explanation:

The interval estimate of a population parameter is an interval of values that consist of the values within which the true value of the parameter lies with a certain probability.

The mean of the sampling distribution of sample proportion is, $\hat p$ .

One of the best interval estimate of population proportion is the 95% confidence interval for proportion,

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Given:

n = 150

$\hat p$ = 0.68

The critical value of <em>z</em> for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.68\pm1.96\sqrt{\frac{0.68(1-0.68)}{150}}\\\\=0.68\pm 0.0747\\\\=(0.6053, 0.7547)\\\\\approx (0.61, 0.75)$

Thus, the reasonable range for the population mean is (61%, 75%).

5 0

3 years ago

Which of the following is a radical equation

Luden [163]

Its b because yeah im just kidding idk im really stupid jk its b hahaha

3 0

2 years ago

Evaluate square root of 16 minus x end root when x = 8.

My name is Ann [436]

$\sqrt{16-x}$
x=8
16-8=8
$\sqrt{8}$

Use a calculator to find the square root of 8.

7 0

3 years ago

Is 3x^2 + x^2 equal to

wariber [46]

The answer is a)3x^4

7 0

2 years ago

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