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Reil [10]
3 years ago
12

An investment account earns 4% per year compounded annually. If the initial investment was $4,000.00, how much is in the account

after 3 years? Round your answer to the nearest dollar.
Mathematics
1 answer:
Alisiya [41]3 years ago
3 0
The account is $4,480 after 3 years with a 4% investment
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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
kap26 [50]

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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Find the value of x. <br> (6x + 1)<br> (9x - 23)
shtirl [24]

Answer:

Step-by-step explanation:

(6x+1)(9x-23)= 6x(9x-23)+1(9x-23)

= 54x²-138+9x-23

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2 years ago
Can someone explain me how to solve this?
sesenic [268]

Answer:

X=12

Step-by-step explanation:

since it is an isosceles triangle opposite sides are congruent which means 7x+24=13x-48 you then would need to subtract either the whole numbers or the variables from each side so add 48 on each side it equals to 7x+72=13x then subtract 7x from 13x and you get 72=6x divide both sides 6/6=x, 72/6 Therefore x=12 HOPE that helped mark me brainiest pls

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Mazyrski [523]

Answer: a and c

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3 years ago
Y + 0.5xy = x + 2; x = 1<br> How do you solve for y
vfiekz [6]

Step-by-step explanation:

replace the value of x with 1

y + 0.5 × 1 × y = 1 + 2

do the maths

y + 0.5y = 3

here the value of y is equal to 1. same goes for x, if we have no given value for them.

1.5y = 3

y = 3/1.5

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