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3 years ago

Evaluate 5x – 2y + (7x – y) for x = 7 and y = –2. 90 –18 –45 63

Mathematics

2 answers:

hram777 [196]3 years ago

5 0

5x - 2y + (7x - y).....x = 7 and y = -2
now we sub
5(7) - 2(-2) + (7(7) -(-2)
35 + 4 + (49 + 2)
35 + 4 + 51
90 <===

Vsevolod [243]3 years ago

5 0

I was going to answer this question because I'm doing the same thing in Maths but it's already answered. He is correct though

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A cuboid is 12 cm broad and 10 cm high. If its volume is 1800 cm³, find its length.

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it's length is 15cm

Step-by-step explanation:

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Answer:

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3 years ago

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steposvetlana [31]

Answer:

y-6 = -7x + 21

Step-by-step explanation:

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3 years ago

(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:

The probability is $P_T= 0.4560$

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

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

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

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

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

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

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

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

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

$=0.001925 -0.000625$

$P_F= 0.0013$

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3 years ago

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