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3 years ago

Hey man. do me a favor and solve . thanx man

Mathematics

1 answer:

yuradex [85]3 years ago

3 0

Answer:

16964.6

Step-by-step explanation:

V=3.14×r2×h r is half remember!

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Can two positive integers have their H.C.F and L.C.M as 12 and 512 respectively? justify.

DaniilM [7]

Here is your answer

Reason:

H.C.F. divides the L.C.M. completely.

Here 512 is not completely divided by 12

So, two numbers can't have their H.C.F and L.C.M as 12 and 512 respectively.

HOPE IT IS USEFUL

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3 years ago

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336 oz equals to ho manylbs

Ann [662]

16 ounces is a pound and the answer is 21 pounds

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4 years ago

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Carol Ann buys a TV at 70% off the original price. If she pays $120, what was the original price of the TV?

Keith_Richards [23]

If the TV is 70% off, then that means she paid 30% of the original cost. If 30% of the original cost is $120, then...

30% * x = 120
.30x = 120
x = 120/.30
x = 400

The original cost was $400

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4 years ago

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In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random

kupik [55]

Answer:

a) $\alpha = 3, \beta = 2$

b) 0.0620

Step-by-step explanation:

We are given the following in the question:

Population mean, $\mu$ = 6

Variance, $\sigma^2$ = 12

a) Value of $\alpha, \beta$

We know that

$\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12$

Dividing the two equations, we get,

$\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3$

b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

We can write the probability density function as:

$f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0$

We have to evaluate:

$P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620$

0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

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4 years ago

What are the characteristics of a gas?

horsena [70]

D. no definite shape or definite volume

7 0

4 years ago

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Hey man. do me a favor and solve . thanx man​

Hey man. do me a favor and solve . thanx man