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Rudiy27
2 years ago
10

Lake Snider High School surveyed its student population about if they prefer pizza or sushi. The students who prefer pizza repre

sent twenty-seven fewer students than five times the number of students who prefer sushi. If 1,528 students prefer pizza, how many students prefer sushi?
Mathematics
1 answer:
stira [4]2 years ago
6 0

Answer:

311 students prefer sushi

Step-by-step explanation:

Let

x------> the number of students who prefer pizza

y -----> the number of students who prefer sushi

we know that

x=5y-27 -----> equation A

x=1,528 -----> equation B

Substitute the equation B in equation A and solve for y

1,528=5y-27\\ 5y=1,528+27\\ 5y=1,555\\ y=311\ students

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Answer:

Option A is correct.

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Step-by-step explanation:

An image of the steps to obtain the equation that relates x to y is attached to this answer.

n = number of variables = 9

μₓ = mean of variable x (rainfall in inches)

μᵧ = mean of variable y (yield in bushels per acre)

σₓ = standard deviation of variable x (rainfall in inches)

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when x = 7.1

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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
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Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

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c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

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