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rjkz [21]
3 years ago
14

which of the following is the solution of -4(5-2x) = 8? (you dont have to give me the answer id like to know how to do it to)

Mathematics
2 answers:
arsen [322]3 years ago
8 0
I'm going to try and make this as simple as possible to explain.

-4(5 - 2x) = 8 

First of all, we have to distribute the outside term (-4) throughout the inside terms on the parenthesis (5 - 2x). Basically what we're doing is multiplying -4 by each term. 

-20 + 8x = 8

Why +8x? It's because a negative times a negative is a positive. 

Now, since we want x all by itself, we have to move the constant, -20, to the other side by adding it on both sides.

8x = 28 

In order to have x all by itself, you have to divide both sides by 8. 
Once you divide 28 by 8, you'll get the quotient and that, ultimately, will be x.

Hope this made it simple and understandable. 
ziro4ka [17]3 years ago
7 0
You are trying to get the x by its self so move the -4 to the other side so that you have 5-2x = 8/-4 then subtract 5 from both sides of the equation -2x = (8/-4) - 5 then once again divid so x = ((8/-4) - 5) / -2 
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Answer:

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Step-by-step explanation:

Let <em>X</em> = number of motherboards that pass the inspection.

The probability that a motherboards pass the inspection is P (X) = <em>p</em> = 0.85.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is,

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that at least 13 of the next 15 motherboards pass inspection as follows:

P (X ≥ 13) = P (X=  13) + P (X = 14) + P (x = 15)

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Thus, the probability that at least 13 of the next 15 motherboards pass inspection is 0.6042.

(b)

Computed the expected number of motherboards should be inspected until a motherboard that passes inspection is found as follows:

Expected\ value=\frac{1}{p}=\frac{1}{0.85}= 1.1764\approx1.18

Thus, on average, 1.18 motherboards should be inspected until a motherboard that passes inspection is found.

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