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Ahat [919]
3 years ago
7

What does 20+25+45=90

Mathematics
2 answers:
stiv31 [10]3 years ago
4 0

Answer:

90=90 :D

am about to get banned :<

Step-by-step explanation:

aleksandr82 [10.1K]3 years ago
4 0

Answer:

90=90

Step-by-step explanation:

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Two numbers multiply to be -7 and add to be -6. What are the numbers?<br><br> Example; 2,3
rewona [7]

<u>Given</u>:

Let the two numbers be x and y.

Two numbers multiply to be -7 and add to be -6.

This can be written in equation as,

xy=-7 and

x+y=-6

<u>Value of the two numbers:</u>

Let us determine the value of the two numbers using substitution method.

Substituting y=-6-x in the equation xy=-7, we get;

x(-6-x)=-7

Simplifying, we get;

       -6x-x^2=-7

   x^2+6x-7=0

(x+7)(x-1)=0

x=-7, 1

Thus, the values of x are x = 1,-7

When x = 1 , the equation x+y=-6 becomes y=-7

When x = -7, the equation x+y=-6 becomes y=1

Therefore, the two numbers are 1 and -7

3 0
3 years ago
6y + 2 + 3 + 14y
hichkok12 [17]

Answer:

20y + 5

Step-by-step explanation:

6y + 2 + 3 + 14y

20y + 2 + 3

20y + 5

Have an amazing day!

PLEASE MARK BRAINLIEST!

3 0
2 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<br> help please please
Ahat [919]

Answer:

1. A

2. C

Step-by-step explanation:

I could be wrong, that's my best guess. Im so sorry if im wrong, have a good day

7 0
3 years ago
How to multiply and divide rational numbers?
lutik1710 [3]

To divide rational numbers, you turn the division problem into a multiplication problem by flipping the second rational number. Then you go ahead and multiply the tops and bottoms together to get your answer. If you can simplify your problem before multiplication, you can go ahead and do so to make your problem easier.

5 0
2 years ago
Read 2 more answers
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