Write the equation of the line passing through the point (3, 5) that is

(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability is $P_T= 0.4560$

b

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

4 0

3 years ago

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Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-8,3), B(-3,3), C(-3,6) an

vichka [17]

I have graphed it for u

6 0

4 years ago

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What passes through the points (-1/2 4) and (1/2 6). show your work

Maslowich

Step-by-step explanation:

The equation of line which passes through these points is determined by:

p(x1,y1)=(-1/2,4) p(x2,y2)=(1/2,6)

$\frac{y - y1}{y2 - y1} = \frac{x - x1}{x2 - x1} \\ \frac{y - 4}{6 - 4} = \frac{x + \frac{ 1}{2} }{ \frac{1}{2} + \frac{1}{2} } \\ \frac{y - 4}{2} = \frac{2x + 1}{2} \\ 2(2x + 1) = 2(y - 4) \\ 4x + 2 = 2y - 8 \\ 4x - 2y + 2 + 8 = 0 \\ 4x - 2y + 10 = 0$

Note:if you need to ask any question please let me know.

5 0

3 years ago

Mr. Jacobs 55 years old and tony is 7 years olds , in how many years will mr. Jacobs 4 times as old as tony ?

V125BC [204]

Call the mystery number of years ' Y ' .

In 'y' years, Mr. Jacobs will be (55+y), and tony will be (7+y).

You said that Mr. Jacobs will then be 4 times as old as tony, so . . .

      55 + y = 4 (7 + y)

Eliminate the parentheses:    55 + y = 28 + 4y

Subtract 28 from each side: 27 + y =   4y

Subtract 'y' from each side: 27     =   3y

Divide each side by 3 :       9       = y

In 9 years, Mr Jacobs will be (55 + 9) = 64.
In 9 years, tony will be (7 + 9) = 16.

64 is 4 times 16, and now that tony has turned 16,
he can start spelling his name with a capital ' T ',
like a regular grown-up.

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3 years ago

Show how to make one addend the next tens number. 15+37=?

4vir4ik [10]

Using long addition to evaluate would give you 52. Adding by using long addition would also give you 52. If you simplify, your answer would still be 52.
=52

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3 years ago

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