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3 years ago

If the sum (3x ^ 2 - 2x + 12) + (x ^ 2 + 8x - 2) is multiplied by 2x ^ 2 what is the result written in standard form ?

Mathematics

1 answer:

alexandr402 [8]3 years ago

5 0

(3x^2 - 2x + 12) + (x^2 + 8x -2)
= (4x^2+6x+10)(2x^2)
= 4x^2(2x^2)+6x(2x^2)+10(2x^2)
= 8x^4 + 12x^3 + 20x^2

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Claim amounts for wind damage to insured homes are independent random variables with common density f(x) = ( 3 x4 , x > 1 0 ,

Irina-Kira [14]

$f_X(x)=\begin{cases}\dfrac3{x^4}&\text{for }x>1\\\\0&\text{otherwise}\end{cases}$

This distribution has expectation

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\int_1^\infty\frac3{x^3}\,\mathrm dx=\frac32$

a. The probability that $X$ falls below the average/expectation is

$P\left(X$

b. Denote by $X_{(3)}$ the largest of the three claims $X_1,X_2,X_3$ . Then the density of this maximum order statistic is

$f_{X_{(3)}}(x)=3f_X(x)F_X(x)^2$

where $F_X(x)=P(X\le x)$ is the distribution function for $X$ . This is given by

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

So we have

$f_{X_{(3)}}(x)=\begin{cases}0&\text{for }x1\end{cases}$

and the expectation is

$E[X_{(3)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(3)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^3}\left(1-\frac1{x^3}\right)^2\,\mathrm dx=\frac{81}{40}=\boxed{2.025}$

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so the expectation is

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lana66690 [7]

Answer: 5,4

Step-by-step explanation:

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3 years ago

In a box of 10 calculators, one is defective. In how many ways can four calculators be selected, if you know one

Juliette [100K]

Given:

Total number of calculators in a box = 10

Defective calculators in the box = 1

To find:

The number of ways in which four calculators be selected and one of the four calculator is defective.

Solution:

We have,

Total calculators = 10

Defective calculators = 1

Then, Non-defective calculator = 10-1 = 9

Out of 4 selected calculators 1 should be defective. So, 3 calculators are selected from 9 non-defective calculators and 1 is selected from the defective calculator.

$\text{Total ways}=^9C_3\times ^1C_1$