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Salsk061 [2.6K]
3 years ago
14

• The graph shows the amount of money Nationwide Shipping Company pays

Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

Step-by-step explanation:

From the graph attached,

1). Amount drivers are paid per mile = \frac{\text{Total amount paid}}{\text{Distance traveled}}

                                                            = \frac{40}{100}

                                                            = $0.40 per mile

   This amount per mile is same at every point of the graph.

   Therefore, the amount drivers are paid per mile is the same for each distance, so the relationship is proportional.

2). Let the relation between the distance traveled (x) and amount paid (y) is,

   y = kx

   k = \frac{y}{x}

   k = \frac{40}{100}

   k = $0.40

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2.-3(x-2) OS<br> (1 Point)<br> Enter your answer
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Answer:

-3x+6

Step-by-step explanation:

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Maya needs $58 to go on a field trip. She has saved $18.50. She earns $6.50 per hour cleaning her neighbor's garden, and she ear
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She has $18.50 so she needs $39.50.

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Answer:

a) 99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b) 99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. 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.

4​% rate of disk drive failure in a year.

This means that 96% work correctly, p = 0.96

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 2

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984

99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b. If copies of all your computer data are stored on four independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

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