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vazorg [7]
3 years ago
8

I need help with this please. Thanks!

Mathematics
1 answer:
den301095 [7]3 years ago
3 0

Answer:

-1

Step-by-step explanation:

2.5 - 3.5x = 8 - 4x - 6

0.5x = 2 - 2.5

0.5x = -0.5

x = -1

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A line passes through the points (0, –3) and (2, –4) . What is the slope of the line?
Fudgin [204]
<span>(0, –3) (2, –4)
slope = difference in y divided by </span><span>difference in x
</span>
<span>slope = (-3 --4) / (0 -2) =
1 / -2

</span>
3 0
3 years ago
Read 2 more answers
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
HELP I NEED HELP ASAP
harina [27]

Maybe the Answer is B.

B.) 0.03%

8 0
3 years ago
Is it a,b,c or d I have 5 minutes please hurry
snow_lady [41]

Answer:

its a

Step-by-step explanation:

4/12 = 1/3

5 0
3 years ago
Read 2 more answers
rick jogged the same distance on tuesday snd friday, and 8 miles on sunday for a total of 20 miles for the week. solve to find t
saveliy_v [14]

Ricky jogged 6 miles on tuesday and 6 miles on friday

<em><u>Solution:</u></em>

Given that,

Rick jogged the same distance on tuesday and friday

Let "x" be the distance jooged on each tuesday and friday

He also jogged for 8 miles on sunday

Total of 20 miles for the week

Therefore, we frame a equation as,

total distance jogged = miles jogged on tuesday + miles jogged on friday + miles jogged on sunday

20 = x + x + 8

20 = 2x + 8

2x = 20 - 8

2x = 12

x = 6

Thus Ricky jogged 6 miles on tuesday and 6 miles on friday

3 0
3 years ago
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