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sveticcg [70]
3 years ago
6

I need help and fast

Mathematics
2 answers:
igomit [66]3 years ago
8 0
It’s “C” I got it right when I had this question before
ch4aika [34]3 years ago
8 0
Ok so first you want to do and then put and that leaves you with
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(5x - 7) + (10x 7) = 180
goldfiish [28.3K]
15x=180
x=12

hope this helps
4 0
3 years ago
Solve: (3w + 4) (2w – 7) = 0
I am Lyosha [343]

Answer:

w=-4/3, 7/2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
If the zeros are 1 and -4, what are the factors?
Natali5045456 [20]

Answer:

(x - 1) and (x + 4)

Step-by-step explanation:

Given the zeros of a polynomial, say x = a, x = b then the factors are

(x - a), (x - b) and the polynomial is the product of the factors, that is

f(x) = (x - a)(x - b)

Here the zeros are x = 1 and x = - 4, thus the factors are

(x - 1) and (x - (- 4)), that is (x - 1) and (x + 4)

5 0
3 years ago
Solve the equation. Show your work.<br>6x-15=3(4x+3)
Anit [1.1K]
6x-15=3(4x+3)
Distribute the three to the parentheses
6x-15=12x+9
Transfer the 6 over to the 12 by subtracting it
-15=6x+9
Subtract 9 from 15
-24=6x
Then divide the six by the -24
X=-4
7 0
3 years ago
Read 2 more answers
6. Assume that a component passes a test is 0.85 and that components perform independently. What is the probability that the thi
Tanzania [10]

Answer:

3.90% probability that the third failure will occur on the tenth component tested

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. 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.

Assume that a component passes a test is 0.85

So they fail with probability of p = 1 - 0.85 = 0.15

What is the probability that the third failure will occur on the tenth component tested

First 9 components: Two failures, that is, P(X = 2) when n = 9.

10th component: Failure with probability 0.15.

So

P = 0.15P(X = 2)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{9,2}.(0.15)^{2}.(0.85)^{7} = 0.2597

So

P = 0.15P(X = 2) = 0.15*0.2597 = 0.0390

3.90% probability that the third failure will occur on the tenth component tested

5 0
3 years ago
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