The following table shows the expressions that represent the sales of two companies:

You own 17 CDs. You want to randomly arrange 5 of them in a CD rack. What is the probability that the rack ends up in alphabetic

amid [387]

The probability that all five end up in alphabetical order is; 1/120.

<h3>What is the probability that the rack ends up in alphabetical order?</h3>

To evaluate the given probability; first, the number of possible arrangements is;

17P5 = 742,560 possible arrangements.

However, the chance of an alphabetical order arrangement in each case is; 1 out of 5! possible arrangements.

Hence, we have that the number of possible alphabetical arrangement is; (1/120) × 742,560 = 6188.

Hence, the required probability is; 6188/742,560 = 1/120

Read more on probability;

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5 0

2 years ago

What is the scale factor of this dilation?<br> 2/3<br> 1 1/2<br> 3<br> 5

Andre45 [30]

$\textbf{Answer:}$

$\textrm{The scale factor of this dilation is }\frac{2}{3}$

$\textbf{Step-by-step explanation:}$

$\textrm{Divide the lengths of triangle A to triangle A prime}$

$\frac{10}{15} =\frac{2}{3}$

$\frac{6}{9} = \frac{2}{3}$

4 0

3 years ago

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2 The triangle and the rectangle shown below have equal perimeters. Find the measures of the side lengths of the triangle. (Figu

Veronika [31]

Answer:

180

Step-by-step explanation:

Subtract the sum of the two angles from 180 degrees. The sum of all the angles of a triangle always equals 180 degrees. Write down the difference you found when subtracting the sum of the two angles from 180 degrees. This is the value of X.When side lengths are given, add them together.

Solve for a missing side using the Pythagorean theorem.

If we know side-angle-side information, solve for the missing side using the Law of Cosines.

5 0

3 years ago

10. Square is a<br>a. Rectangle<br>b. Rhombus<br>c. Both<br>d. None of these

nikitadnepr [17]

Answer:

d. None of these

Hope it helps

Have a great day : )

4 0

3 years ago

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LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last season. During the summer, h

aleksandr82 [10.1K]

Answer:

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$

$p_v =P(z>2.905)=0.0018$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of free throws made is higher than 0.4

Step-by-step explanation:

Data given and notation

n=40 represent the random sample taken

X=25 represent the number of free throws made

$\hat p=\frac{25}{40}=0.625$ estimated proportion of free throws made

$p_o=0.4$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.4:

Null hypothesis: $p\leq 0.4$

Alternative hypothesis: $p > 0.4$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{40}}}=2.905$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.905)=0.0018$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of free throws made is higher than 0.4

8 0

3 years ago