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lesantik [10]
3 years ago
10

A car is designed to last an average of 12 years with a standard deviation of 0.8 years. What is the probability that a car will

last less than 10 years?
Mathematics
1 answer:
aliya0001 [1]3 years ago
6 0
The probability that the car will last less than 10 years would be .0062. With the given data (M = 12, SD = .8), i computed the z-score and looked it up on the z table. The answer yields .0062 or 0.62%. Thus, the probability is very little.
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C(squared)L(squared) + 100v(squared) = 100c(squared)<br>I have to solve for L
jeyben [28]
First subtract 100v² from both sides to get:

C²L²=100c²-100v²

Then divide both sides by C²:

L²=(100c²-100v²)/C²

Then take the square root of both sides:

L=+ or - the square root of (100c²-100v²)/C²
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2 years ago
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Given this table of values, what is the value of f(-4.5) ? <br><br> help pelase !!
True [87]

Answer:

12.6

Step-by-step explanation:

f(-4.5) means "what is the y-value when x equals -4.5?"

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5c=d for c<br> I need help with answering this math problem
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Step-by-step explanation:

Let's solve for c.

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Solve the system by substitution.<br> x – 4y = -8<br> 5у – 1 = x<br> Submit Answer
Vikki [24]

Answer:

y = -7 and x = -36

Step-by-step explanation:

x - 4y = -8

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→ Simplify

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LUCKY_DIMON [66]
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u(x,y)=xy
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\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}

and has a determinant of

\det\mathbf J=-\dfrac{2x}y

Note that we need to use the Jacobian in the other direction; that is, we've computed

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}

but we need the Jacobian determinant for the reverse transformation (from (x,y) to (u,v). To do this, notice that

\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to

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