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klasskru [66]
3 years ago
12

The life of light bulbs is distributed normally. the standard deviation of the lifetime is 25 hours and the mean lifetime of a b

ulb is 590 hours. find the probability of a bulb lasting for at most 622 hours. round your answer to four decimal places.
Mathematics
1 answer:
Dominik [7]3 years ago
4 0
The probability is 0.8997.

We will use a z-score to answer this question.  z-scores are given by the formula
z=\frac{X-\mu}{\sigma}

With our information, we have
z=\frac{622-590}{25}=\frac{32}{25}=1.28

Looking this up in a z-table (http://www.z-table.com) we see that the area to the left of this (everything less than, up to this value) is 0.8997.
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Does anyone know this?
barxatty [35]

Answer:

\Huge \boxed{-2}

Step-by-step explanation:

f(x)=-2x^2-2x+10

For f(2), the input of the function is 2.

Let x = 2.

f(2)=-2(2)^2-2(2)+10

Evaluate.

f(2)=-2(4)-2(2)+10

f(2)=-8-4+10

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8 0
3 years ago
Read 2 more answers
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amid [387]

Answer: QE = 10

Step-by-step explanation: To solve this problem, it's important to understand that the diagonals of a parallelogram bisect each other.

This means that E is the midpoint of diagonal SQ.

So we can setup the equation x² + 9x = 4x + 6.

To solve this polynomial equation, set it equal to zero first.

So we have x² + 5x - 6 = 0 and we get (x + 6)(x - 1) = 0

when we factor the left side of the equation.

So this means that x = -6 or x = 1.

However, -6 will give us a negative length when we plug it in

to find QE so this will not work.

However, plugging 1 in will give us 10 as a length so QE = 10.

4 0
2 years ago
If y=(sec^-1x)^2, then show that x^2(x^2-1) d^2y/dx^2 + (2x^3 - x) dy/dx = 2
Arisa [49]

Step-by-step explanation:

Hope it helps you in your learning process.

8 0
3 years ago
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ella [17]

Answer:

5 times 3

Step-by-step explanation:

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5 0
3 years ago
What equation can be used to find the length of AC
Leno4ka [110]
I'm pretty sure this is the answer:
(10)sin(40°) = AC

It's <em>sin </em>becasue it <em>needs </em>the opposite side of 40° and <em>has </em>the hypotenuse. 
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3 years ago
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