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3 years ago

How do you find x? Please help

Mathematics

1 answer:

marusya05 [52]3 years ago

5 0

Since you know that DA and DC are equal in length, you can say 2x-3=x+2, when you add 3 to both sides, it becomes 2x=x+5, subtract x from both sides, and you find that x=5

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Arlecino [84]

The listed ordered pairs that are solutions are D

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3 years ago

Is question a. Linear or non linear

Zanzabum

Answer:

yes

Step-by-step explanation:

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3 years ago

help ASAP find the coordinates of the midpoint of the given segment given its endpoints are (-4,3) and (2,-1)

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

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3 years ago

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of

liq [111]

Answer:

The probability of failure of both the bulbs is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

$f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx$

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

$P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx$

Similarly for the bulb Y the distribution function is given by

$P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy$

Thus the probability for both the bulbs to fail within 1500 hours is

$P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323$

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3 years ago

What is the value of g?

lyudmila [28]

Answer:

Step-by-step explanation:

I'm not sure about the answer

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2 years ago

Read 2 more answers