Please explain this is math homework

In Mrs. Hu’s classroom, 4/5 of the students have a dog as a pet. Of the students who have a dog as a pet, 2/3 have a cat as a pe

tresset_1 [31]

2/3 of the 4/5 of the class has a dog and a cat. Just multiply
(2/3)(4/5)=8/15
45*(8/15)=24

6 0

3 years ago

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Suppose it is known that the distribution of purchase amounts by customers entering a popular retail store is approximately norm

Ket [755]

Answer:

69.15% probability that a randomly selected customer spends less than $105 at this store

Step-by-step explanation:

Problems of normally distributed samples are 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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 100, \sigma = 10$

What is the probability that a randomly selected customer spends less than $105 at this store?

This is the pvalue of Z when X = 105. So

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

$Z = \frac{105 - 100}{10}$

$Z = 0.5$

$Z = 0.5$ has a pvalue of 0.6915

69.15% probability that a randomly selected customer spends less than $105 at this store

7 0

3 years ago

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There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$

The z value is calculated as before:

$z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0$

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0

3 years ago

Ron practiced piano 1 2/3 hours on monday and 5/6 hour on tuesday. how much did he practice in all those two days

insens350 [35]

2.5 hours i think but i might be wrong :)

4 0

3 years ago

The radius of a circle is 42 inches. What is the circle's area?<br> Use 3.14 for .

Mazyrski [523]

Answer:

A =5538.96 in^2

Step-by-step explanation:

The area of a circle is given by

A = pi r^2

We know the radius is 42 and pi is approximated by 3.14

A = 3.14 * 42^2

A =5538.96 in^2

3 0

3 years ago

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