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Sever21 [200]
2 years ago
8

11. Joe bought 3 pairs of jeans for $71.40. a. How much did he pay per pair? b. How much would be need to pay for 8 pairs of jea

ns? C. Solve by setting up a proportion​
Mathematics
2 answers:
vodomira [7]2 years ago
5 0

Answer:

a. $23.80 per pair

b. $190.40

Step-by-step explanation:

3 pairs of jeans for $71.40

so the proportion is 71.40:3 = 23.8:1

23.8:1 = 8(23.8):8(1) = 190.4:8

Vikki [24]2 years ago
4 0

Answer:

$23.80 per pair, $190.4 for 8 pairs

Step-by-step explanation:

a. assuming each pair of jeans is the same price, just divide the total by 3. the price per pair is 71.40/3, which is $23.80 per pair

b. simply multiply the price of one pair of jeans, as found above, by 8. 23.80*8 is $190.4 for 8 jeans

hope this helped!

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12x12=144. Answer is 144 miles

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Answer:

243 and 2 over 5

Multiply

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3 years ago
A certain brand of golf balls comes in packs of 24. Each pack has three yellow golf balls in it. Willam orders 360 golf balls on
iogann1982 [59]

Start by taking the total number of golf balls, 360, and dividing it by how many golf balls come in each pack, to find the number of packs.


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4 years ago
Read 2 more answers
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
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