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3 years ago

Can someone help I know it's easy. ( picture attached)

Mathematics

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

B 125.6

Step-by-step explanation:

We can use the formula for the circumference to solve.

C=2 $\pi$ r

C=2 $\pi$ (20)

C=(6.28)(20)

C=125.6

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Answer:

$319

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3 years ago

Equivalent expression for 90x+120

NNADVOKAT [17]

90x+120 is equivalent to 30 (3x+4)

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3 years ago

What kind of proportion tells that when one quantity is increasing or decreasing, the other quantity also increasing or decreasi

Zanzabum

Answer:

Step-by-step explanation:

Direct proportion is when two numbers move in the same direction. If one of the numbers increase, the other number increases and if the one of the numbers decrease, the other number decreases. The magnitude of the movement of the numbers is bounded by a constant

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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 best way to display data if there is a wide range and we want to see individual data?

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3 years ago

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