It takes half a yard of ribbon to make a bow. how many bows can be made with 7 yards of ribbon?

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

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

So

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

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

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

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

3 years ago

The histogram below shows the number of weeks Jimmy spent different amounts of money to fill the gas tank in his car. If the gra

sammy [17]

Answer:

A. The difference in the heights of the bars will not appear as great.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Write an equation to represent the following statement. 24 is 4 times as great as k.

Artyom0805 [142]

Answer:

24 x 4 = k x 4

Step-by-step explanation:

8 0

3 years ago

A degree is larger than 30 arc minutes. ...? true or false

oksano4ka [1.4K]

It is a completely true statement that a degree is larger than 30 arc minutes. The correct option among the two options that are given in the question is the first option. 30 arc minutes is almost equal to 1/2 degree. I hope that this is the answer that you were looking for and the answer has actually come to your great help.

7 0

3 years ago

Read 2 more answers

A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.

oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

$H_0: \mu=2.4\\\\H_a:\mu> 2.4$

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58$

The degrees of freedom for this sample size are:

$df=n-1=45-1=44$

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.5801)=0.0066$

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0

3 years ago

It takes half a yard of ribbon to make a bow. how many bows can be made with 7 yards of ribbon?​

It takes half a yard of ribbon to make a bow. how many bows can be made with 7 yards of ribbon?