All categories

3 years ago

what is the diameter of 1e-5 meters? PLEASE GIVE ME THE RIGHT ANSWER I WILL MARK U AS BRAINLIEST... :)

1 answer:

4 0

I'm not sure exactly what you're asking but 1×10^-5 is .00001. I could be interpreting your question wrong in which case I'm sorry.

You might be interested in

David rents DVDS from a company that changes a $7.00 monthly fee and $1.50 for each dvd rental. He ends up paying $20.50 for the

serious [3.7K]

1.5x + 7 = 20.5
First subtract 7 from both sides
1.5x = 13.5
Then divide both sides by 1.5
x = 9

Let me know in comments if you have questions

3 0

3 years ago

Which 1 weighs about 6 ounces a toothbrush or a paper clip?

bogdanovich [222]

A toothbrush weighs that much. a paperclip weighs less then an ounce, about 0.049 oz.

have a nice day

7 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 $n = 4$