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Vedmedyk [2.9K]
3 years ago
9

Vivian divides the total amount of money she has evenly between her three children. Each child receives $8.84. Let m be the tota

l money that Vivian has. Write an equation involving m and solve it for the total amount of money that Vivian has (or had).
Mathematics
2 answers:
Flura [38]3 years ago
5 0

Answer:

m/3=8.84

3(m/3)=3(8.84)

m= $26.52

Step-by-step explanation:

salantis [7]3 years ago
5 0
M=26.52
Hope it helped :)
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A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li
svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

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}.\pi^{x}.(1-\pi)^{n-x}

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

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

And \pi is the probability of X happening.

In this problem

15 drivers are randomly selected, so n = 15.

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year.  This means that \pi = 0.08.

What is the probability of getting 3 or more who were involved in a car accident last year?

This is P(X \geq 3).

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863

P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734

P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887

Finally

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

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Answer:

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