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grandymaker
2 months ago
5

Mr. Pratt has 26 students in his math class. He has three prizes to give away: a pencil, an eraser, and a homework pass. How man

y ways can he choose three students to win these awards?
Mathematics
1 answer:
lawyer [12.5K]2 months ago
3 0

Answer:

Mr. Pratt can award three students in 2,600 different ways.

Step-by-step explanation:

The method of choosing k items from n unique items, without any repetitions, is termed as combinations in mathematical terms.

The formula to calculate the number of combinations for k items from n is:

{n\choose k}=\frac{n!}{k!(n-k)!}

Mr. Pratt has three awards to distribute.

The total number of students present in his classroom is n = 26.

He has k = 3 awards, specifically a pencil, an eraser, and a homework pass.

To find out how many combinations are possible for Mr. Pratt to choose three students from 26 to receive the awards, we calculate as follows:

{n\choose k}=\frac{n!}{k!(n-k)!}

     =\frac{26!}{3!(26-3)!}

     =\frac{26!}{3!\times 23!}

     =\frac{26\times25\times24\times23!}{3!\times23!}

     =2600

Therefore, Mr. Pratt can choose three students to win these awards in 2,600 different ways.

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Step-by-step explanation:

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2 months ago
If y varies directly as x, and y is 400 when x is r and y is r when x is 4, what is the numeric constant of variation in this
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Answer:

10

Step-by-step explanation:

If there’s a direct correlation between variables x and y, it can be expressed as

y = kx,

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_____________________________

For the first scenario

y = 400

x = r

using y = kx, we can derive the relation:

400 = kr

To find k here:

k = 400/r

For the second scenario

y = r

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again using y = kx, the relationship is given by:

r = 4k

Solving for k:

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Since both conditions provide the same equation, k will have the same value in both cases.

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Hence, r is calculated to be 40.

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The constant of variation is confirmed to be 10, which is the correct answer.

Verification:

k = 400/r = 400/40 = 10

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