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Gennadij [26K]
3 years ago
15

You want to save for a brand-new car. You put the $5,000 your Grandma gave you when you graduated in an account that pays 6% int

erest and is compounded monthly. How much will you have at the end of five years?
Mathematics
1 answer:
baherus [9]3 years ago
5 0

Answer:

Balance at the end of 5 years: $7,346.64

Step-by-step explanation:

A = P(1+r/n)nt

Principal: 5,000

Rate of Interest: 6%

Number of Times, Compounded In A Year: 12

Year: 5 years

A = 5,000(1+1.06/12)^5

A = 5,000(1.08)^5

A = 7346.64

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Assume that there is a 4​% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit
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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

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

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?

This is P(X \ geq 1) when n = 4

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

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

P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744

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

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How does the graph of y=-3√2x-4 compare to the graph of y=-3√x-4?
kifflom [539]
<span>Assuming the graph is y=-3(√2x)-4 and y=-3√(x-4) the transformation would be:

</span><span>The graph is compressed horizontally by a factor of 2
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Answer: C. <span>The graph is compressed horizontally by a factor of 2, moved left 4, and moved down 4.
</span>
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If 1 inch equals 20miles then 8 inches equals 20 x 8 = 160 miles.

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