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3 years ago

12. The results of an election for mayor are shown

Mathematics

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

Assuming the group of the candidates is ordered the same way as the votes they received, Margaret O'Conner received the MOST votes, at 12,926, while Leonard Hansen received the LEAST votes, at 12,409.

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(square root) -28
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Albert jumped 100 times in 97.5 seconds. Round off the time taken to the nearest second.

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3 years ago

A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

Eddi Din [679]

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

$D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{1}{455} = 0.0021$

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

$D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36$

Total outcomes:

12 peaches, from a set of 15. So

$T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455$

Probability:

$p = \frac{D}{T} = \frac{36}{455} = 0.0791$

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

3 0

3 years ago