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Ad libitum [116K]
3 years ago
14

The sum of z and 7 algebraic expression

Mathematics
1 answer:
snow_tiger [21]3 years ago
6 0

Step-by-step explanation:

The algebraic expression for The sum of z and 7 algebraic expression is given as:

z + 7

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Pls help me on this
LuckyWell [14K]

Answer:

y = 2(x-3)

Step-by-step explanation:

This is because Gavin representing "x" loses 3 marbles, which is shown by the -3 in the parenthesis, is half of what Cindy has, or as the problem says "Cindy has double the marbles that Gavin has." These two together result in y=2(x-3)

3 0
3 years ago
Jamie can work no more than 20 hours a week this summer. She needs to earn at least $256 a week to save enough
ivanzaharov [21]
She will have to get 12.8 per hour
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3 years ago
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

7 0
3 years ago
Find an equation of the parabola that has the indicated vertex and whose graph passes through the given point
vichka [17]
Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1

Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2

Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3
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3 years ago
What is 8x -5+2 x simplified
IgorC [24]

8x -5+2 x

8x+2x-5 ( combine like terms)

answer:

10x-5 or -5+10x

8 0
4 years ago
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