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2 years ago

A right rectangular prism has a length of 7 centimeters, a width of 5 centimeters, and a height of 4 centimeters

Mathematics

1 answer:

Julli [10]2 years ago

8 0

Answer:

The volume is 140

Step-by-step explanation:

V = l × w × h

V = 7 × 5 × 4

V = 140

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Fittoniya [83]

1/3 that would be it

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3 years ago

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1. Place a _____________ on the number line for the number in the expression.

nevsk [136]

Answer:

the answer is a equation

Step-by-step explanation:

because u supposed to make it a equation in order for it to be a expression.

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3 years ago

What can you conclude about the signs of two integers whos product i a positive and negative

KengaRu [80]

If two integers product is:

Positive: then the integers were both negative or both positive (same sign)
Negative: then the integers consisted of one negative and one positive (opposite signs)

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3 years ago

Find f^-1(x) where f(x) = 3x + 2 / x

dimulka [17.4K]

Answer:

denoted interchange solve

Step-by-step explanation:

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3 0

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

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

6 0

3 years ago