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2 years ago

The value of y varies directly with x. When y = 3.5, x = 2. What is the value of y when x is 16?

1 answer:

4 0

In order to do this problem, you have to divide 16 by 2 to see how many times x was multiplied. you would find that x was multiplied 8 times, so in order to find y, you would multiply 3.5 eight times.

y = 28 when x = 16

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3 years ago

3÷3912.00 carry to the hundreth

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2 years ago

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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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3 years ago

Reduce this algebraic fraction. y^3 -3y^2+y-3 / y^2-9

blondinia [14]

Answer:

y³ - 3y² + 1/y+3

Step-by-step explanation:

y² - 9 = (y - 3) ( y + 3)

y-3/(y-3)(y+3) = 1/y+3

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2 years ago

Read 2 more answers

Suppose you had D dollars in your bank account. you spent $22 but have at least $28 left. how much money did you have initially?

nalin [4]

Answer:

Answer:

d--22→ 28

d→ 50

Step-by-step explanation:

Let d dollars be the initial amount in your bank account.

We have been given that you spent $22, so the amount left after spending $22 will be: .

We are also told that after spending $22, you have at least $28. This means that the amount left after spending $22 will be greater than or equal to $28.

We can represent this information in an equation as:

Therefore, the inequality represents the initial amount of money you had.

Now let us solve for d by adding 22 to both sides of our inequality.

Therefore, initially you had at least $50.

Still stuck? Get 1-on-1 help from an expert tutor now.

Step-by-step explanation:

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2 years ago