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Maksim231197 [3]
3 years ago
7

5 rate?? my brother doesn't want to helpx me someone help?]'

Mathematics
2 answers:
Luden [163]3 years ago
7 0

Answer:

-5/2 is the reciprocal of -2/5, but the opposite would just be 5/2.

Step-by-step explanation:

S_A_V [24]3 years ago
6 0

Answer:

yes the answer is yes i think im not good at explaing but hope that helps

Step-by-step explanation:

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What is the equation of the line that passes through the point (3,6) and has a slope of 4/3
Neko [114]

Answer:

y = 4/3x+2

Step-by-step explanation:

We can use the slope intercept form of the equation

y = mx+b

Where m is the slope and b is the y intercept

y= 4/3 x +b

Substitute the point into the equation

6 = 4/3(3) +b

6 = 4 +b

Subtract 4 from each side

2 = b

y = 4/3x+2

7 0
3 years ago
Idk this bs its stupid.?!
LekaFEV [45]

Answer:y=-3/2x-1

Step-by-step explanation:-3/2 is your slope which is your m and -1 is your b x is left alone.

7 0
4 years ago
Read 2 more answers
Solve for x <br><br> 3(x+2)+4(x-5)=10
Natasha_Volkova [10]

Answer:

x=24/7

Step-by-step explanation:

3(x+2)+4(x-5)=10

3x+6+4(x-5)=10

3x+6+4x-20=10

7x+6-20=10

7x-14=20

7x=24

x=24/7

8 0
3 years ago
What is the equation of the line that passes through the point (-5, 7)<br> and has a slope of –1/5?
Snowcat [4.5K]
The equation is y = -1/5x + 6
8 0
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
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