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marta [7]
3 years ago
12

A high school soccer team is going to Columbus to see a professional soccer game. A coordinate grid is superimposed on a highway

map of Ohio. The high school is at point (2, 4) and the stadium in Columbus is at point (7, 6). One unit on the map is 3 miles. The approximate distance between the high school and the stadium to the nearest mile is Answer miles.
Mathematics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

16 miles

Step-by-step explanation:

Given that :

1 unit = 3 miles

High School = (2, 4)

Stadium = (7, 6)

Approximate distance between high school and stadium :

Distance between High school and Stadium

x1 = 2 ; y1= 4 ; x2 = 7 ; y2 = 6

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Distance = sqrt((7 - 2)^2 + (6 - 4)^2)

Distance = sqrt((5^2 + 2^2))

Distance = sqrt(25 + 4)

Distance =5.3851648 units

1 unit = 3 miles

Approximate Distance in miles :

3 * 5.3851648 = 16.155494

Approximate distance = 16 miles

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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
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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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