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joja [24]
3 years ago
15

How many different ways could a baking contest be judged if 15 pies are entered and 4 ribbons are awarded?

Mathematics
2 answers:
Rashid [163]3 years ago
8 0
This is a question of combination since it does not take into account the order of the pies. The answer would be 1365. There are 1365 ways a baking contest ways can be judged if 4 ribbons are awarded with 15 pie entries. 

In case you need the order, then the permutation answer would be 32760.
RoseWind [281]3 years ago
7 0

Answer:

1365

Step-by-step explanation:

We are given that 15 pies are entered in the contest.

Out of 15 , 4 are awarded with ribbons.

Now we are supposed to find How many different ways could a baking contest be judged.

Since the order of pie doesn't matter over here.

So, we will use combination.

^nC_r=\frac{n!}{r!(n-r)!}

Substitute n = 15

r = 4

So, ^{15}C_4=\frac{15!}{4!(15-4)!}

^{15}C_4=\frac{15!}{4!(11)!}

^{15}C_4=\frac{15 \times 14\times 13 \times 12 \times 11!}{4!(11)!}

^{15}C_4=\frac{15\times 14\times 13 \times 12}{4\times 3 \times 2\times 1}

^{15}C_4=1365

Hence there are 1365 ways in which a baking contest could be judged.

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Answer:

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Step-by-step explanation:

First ball:

Probability of drawing a white ball is 5/8

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Second ball:

This depends on the first ball drawn, lets say you drew a white ball initially, 4 white balls are left out of 7 balls in total. The probability of a white ball in the second pick is 4/7.

Total probability of drawing two white balls is 5/8*4/7 (since they are independent events).

If you picked a black ball initially, picking another black ball would have a probability of 2/7, on similar grounds , total prob for 2 blacks would be 3/8*2/7.

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Answer:

\textsf{A)} \quad 5u^3-2u^2+5u+8

Step-by-step explanation:

Given expression:

(-2u^3+5u-1)+(7u^3-2u^2+9)

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Answer:

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Step-by-step explanation:

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