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3 years ago

11

Chance knows the length of a football field (including end zones) is 120 yards. The width of the football field is 53.33 yards.

What is the area of the football field, rounded appropriately?

2 answers:

Ira Lisetskai [31]3 years ago

6 0

<span>6399.6 yds </span>approximately

Sphinxa [80]3 years ago

6 0

The answer is 6,399.6 but if you round it up it would be 6,400.

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In the united states, voters who are neither democrat nor republican are called independent. it is believed that 12% of voters a

Nitella [24]

Answer:

a) 0.0245 = 2.45%

b) 0.9483 = 94.83%

c) 0.6931 = 69.31%

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are independent, or they are not. The probability of a person being independent is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

It is believed that 12% of voters are independent.

This means that $p = 0.12$

Survey of 29 people

This means that $n = 29$

a. what is the probability that none of the people are independent? probability =

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

0.0245 = 2.45% probability that none of the people are independent

b. what is the probability that fewer than 7 are independent? probability =

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

$P(X = 1) = C_{29,1}.(0.12)^{1}.(0.88)^{28} = 0.0971$

$P(X = 2) = C_{29,2}.(0.12)^{2}.(0.88)^{27} = 0.1853$

$P(X = 3) = C_{29,3}.(0.12)^{3}.(0.88)^{26} = 0.2274$

$P(X = 4) = C_{29,4}.(0.12)^{4}.(0.88)^{25} = 0.2016$

$P(X = 5) = C_{29,5}.(0.12)^{5}.(0.88)^{24} = 0.1374$

$P(X = 6) = C_{29,6}.(0.12)^{6}.(0.88)^{23} = 0.0750$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0245 + 0.0971 + 0.1853 + 0.2274 + 0.2016 + 0.1374 + 0.0750 = 0.9483$

0.9483 = 94.83% probability that fewer than 7 are independent

c. what is the probability that more than 2 people are independent? probability =

Either two or less are, or at least 3 are independent. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2) + P(X > 2) = 1$

We want P(X > 2). So

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

$P(X = 1) = C_{29,1}.(0.12)^{1}.(0.88)^{28} = 0.0971$

$P(X = 2) = C_{29,2}.(0.12)^{2}.(0.88)^{27} = 0.1853$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0245 + 0.0971 + 0.1853 = 0.3069$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.3069 = 0.6931$

0.6931 = 69.31% probability that more than 2 people are independent

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3 years ago

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3 years ago

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