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Drupady [299]
3 years ago
5

A family is planning a three-week vacation for which they will drive across the country. They have a van that gets 12 miles per

gallon, and they have a sedan that gets 36 miles per gallon. How much more will they pay for gasoline if they take the van? (Assume that the family will drive 2500 miles and that gas costs $2.50 a gallon. Round your answers to the nearest cent.)
Mathematics
1 answer:
svetoff [14.1K]3 years ago
8 0

Answer:

The family will pay 34722 cents more if they take the van

Step-by-step explanation:

Given

Van = 12 miles per gallon

Sedan = 36 miles per gallon

Distance = 2500 miles

Gas = $2.50 per gallon

First, we need to determine the number of gallons that'll be used by both vehicles

This  is done by dividing total distance by number of miles per gallon

Van = \frac{2500}{12} \ gallon

Sedan = \frac{2500}{36}\ gallon

Next, is to multiply this by the cost of gas per gallon;

This gives the total spendable amount on both vehicles

Van = \frac{2500}{12} * \$2.50

Van = \frac{\$6250}{12}

Van = \$520.833

Sedan = \frac{2500}{36}* \$2.50

Sedan = \frac{\$6250}{36}

Sedan = \$173.611

Next is to get the difference between these amounts

Difference = \$520.833 - \$173.611

Difference = \$347.222

Multiply by 100 to convert to cents

Difference = 347.222 * 100 cents

Difference = 34722.2 \ cents

Difference = 34722\ cents (Approximated)

Hence;

<em>The family will pay 34722 cents more if they take the van</em>

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A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters
natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used 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.

70% of all eligible voters will vote in the next presidential election.

This means that p = 0.7

20 eligible voters were randomly selected from the population of all eligible voters.

This means that n = 20

What is the probability that fewer than 11 of them will vote?

This is:

P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)

In which

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

P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308

P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120

P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039

P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010

P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002

P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479

0.0479 = 4.79% probability that fewer than 11 of them will vote

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