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4 years ago

14

Jen uses 8.6 gallons of gasoline to drive 164.7 miles. Which is the best estimate for thenumber of miles per gallon of gasoline

her vehicle gets?
A. 10 miles per gallon

B. 16 miles per gallon

C. 20 miles per gallon

D. 30 miles per gallon

1 answer:

zimovet [89]4 years ago

4 0

Answer:

C

Step-by-step explanation:

164.7/8.6 = 19.15

about 20 miles per gallon

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A manager randomly selected 25 large cash transactions at a bank that were made in January. The manager then carefully tracked d

Dmitriy789 [7]

Answer:

0.4629 = 46.29% probability that the manager finds more than two such transactions

Step-by-step explanation:

For each transaction, there are only two possible outcomes. Either there is a procedural error, or there is not. The probability of a procedural error on a transaction is independent of any other transaction. This means that 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.

A manager randomly selected 25 large cash transactions at a bank that were made in January.

This means that $n = 25$

The chance for a procedural error is 10%

This means that $p = 0.1$

What is the probability that the manager finds more than two such transactions?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{25,0}.(0.1)^{0}.(0.9)^{25} = 0.0718$

$P(X = 1) = C_{25,1}.(0.1)^{1}.(0.9)^{24} = 0.1994$

$P(X = 2) = C_{25,2}.(0.1)^{2}.(0.9)^{23} = 0.2659$

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0718 + 0.1994 + 0.2659 = 0.5371$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5371 = 0.4629$

0.4629 = 46.29% probability that the manager finds more than two such transactions

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