Pls help pls i need help

0.05 is 1/10 the value of A. 5 B. 0.005 C. 0.5 D. 50

anastassius [24]

Answer:

C.

Step-by-step explanation:

as 0.05 is 1/10 of a number, multiplying 0.05 by 10 will give you the number that when divided by 10 will give you 0.05. Hence just multiply 0.05 by 10 and you get 0.5.

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

A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t

Sati [7]

Answer:

$\frac{2}{7}$

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = $\frac{1}{7}$

Remaining fuel after driving around the city = 1 - $\frac{1}{7}$

= $\frac{7 - 1}{7} = \frac{6}{7}$

According to question:

As from the rest of the gas in the car that is $\frac{6}{7}$ , the car can complete 3 trip to Ottawa which means,

By unitary method:

The car can complete 3 trip by using = $\frac{6}{7}$ tank of gas.

The car can complete 1 trip by using = $\frac{6}{7} \div 3$

= $\frac{6}{7} \times\frac{1}{3}$

= $\frac{6}{21}$

= $\frac{2}{7}$ tank of gas

Thus, $\frac{2}{7}$ tank of gas used for each complete trip to Ottawa.

5 0

3 years ago

3 is 12% of what number

AleksAgata [21]

3/12% = 25

Therefore, 3 is 12% of 25.

3 0

4 years ago

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6. The Grand Theater has 13,451 seats. If 15,340 people need to be seated in the theater for a music concert, write and solve

Elena L [17]

Hey There!

To find how many seats are needed to be added, you need to subtract 15,340 by 13,451. You need to find the difference between the seats you need and the seats you have. So...

15,340-13,451= 1,889

You need to add 1,889 more seats.

Hope This helped!

Godspeed,

SongBird

8 0

4 years ago

Read 2 more answers