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Harman [31]
2 years ago
5

What is 7981 km and 943 m in milametres?

Mathematics
2 answers:
s2008m [1.1K]2 years ago
7 0
<span>7 981 943 000 millimeters</span>
lakkis [162]2 years ago
5 0
7 981 943 000 Millimeters
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HURRY! What is the value of x in the diagram? Please leave an explanation fr the answer
melomori [17]
X = -15 because first u move the ax’s to one side so it would be 3x-5x=25+5 then u calculate each side 3x-5x= -2x and 25+5=30 so -2x=30 and then u divide both sides by -2 and -2/-2 cancels out and 30/-2 equals -15 (I hope his helps) :))
6 0
2 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
Evaluate -x +(-5.8) for x= 7.1.
creativ13 [48]
-7.1+-5.8=-12.9
hope this helps :)
7 0
3 years ago
Read 2 more answers
A meter stick perpendicular to the ground casts a 1.5 meter shadow at the same time a telephone pole casts a shadow that is 9 me
snow_lady [41]
The height of the telephone pole = 9 / 1.5 = 6 meters
7 0
3 years ago
Which of the honey jars is the best value for money? Please help ten points :)
KatRina [158]

Answer:

The large jar is best value for money.

Step-by-step explanation:

For large jar you get 540/4.10 = 131.7g of honey per penny.

For small jar you get 360/2.81 = 128.1g of honey per penny

8 0
3 years ago
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