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2 years ago

WORK OUT (810^-3)(2*10^-4)

2 answers:

6 0

(8 x 10^-3) x (2 x 10^-4)

10^-3 = 1/10^3, which is equal to 1/1000

10^-4 = 1/10^4, which is equal to 1/10000

8 x 1/1000 = 8/1000

2 x 1/10000 = 2/10000

(I’m turning these into decimals because it’s easier lol)

0.008 x 0.0002 = 0.0000016

In scientific notation, it is 16 x 10^-7

Hope this helped!

frutty [35]2 years ago

4 0

-First we have to work out what’s in parenthesis.
(8*10^-3)=(8*0.001)=0.008

-This is for the second parenthesis
(2*10^-4)=(2*0.0001)=0.0002

-Now we multiply them together

ANSWER: 0.0000016

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A random variable X with a probability density function () = {^-x > 0

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The solutions to the questions are

The probability that X is between 2 and 4 is 0.314
The probability that X exceeds 3 is 0.199
The expected value of X is 2
The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

$P(x) = \int\limits^a_b {f(x) \, dx$

So, we have:

$P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2$

Expand the expression

$P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}$

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P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

$P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx$

Using an integral calculator, we have:

$P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3$

Expand the expression

$P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}$

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

$E(x) = \int\limits^a_b {x * f(x) \, dx$

So, we have:

$E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx$

This gives

$E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx$

Using an integral calculator, we have:

$E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}$

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

$E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx$

This gives

$E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx$

Using an integral calculator, we have:

$E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0$

Expand the expression

$E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}$

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

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<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

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Hich ordered pair is the solution to the system of equations? {x=(1)/(2)y+5} {2x+3y=-14}

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Answer:

(2, - 6 )

Step-by-step explanation:

Given the 2 equations

x = $\frac{1}{2}$ y + 5 → (1)

2x + 3y = - 14 → (2)

Substitute x = $\frac{1}{2}$ y + 5 into (2)

2( $\frac{1}{2}$ y + 5 ) + 3y = - 14 ← distribute and simplify left side

y + 10 + 3y = - 14

4y + 10 = - 14 ( subtract 10 from both sides )

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Substitute y = - 6 into (1) for corresponding value of x

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3 years ago

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WORK OUT (8*10^-3)*(2*10^-4)

WORK OUT (810^-3)(2*10^-4)