<img src="https://tex.z-dn.net/?f=z%20%7B%7D%5E%7B%20-%204%7D%20y%20%7B%7D%5E%7B%20-%202%7D%20z%20%7B%7D%5E%7B4%7D%20" id="TexFo

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2 years ago

rmula1" title="z {}^{ - 4} y {}^{ - 2} z {}^{4} " alt="z {}^{ - 4} y {}^{ - 2} z {}^{4} " align="absmiddle" class="latex-formula">
l need help solving this

Mathematics

1 answer:

natka813 [3]2 years ago

6 0

.. i just went off of it

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Giovanna and Janet have jobs where they get paid by the hour .Below are the numbers of hours of each have worked on each of the

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I would need more information to help you. Like the numbers of hours worked, amount of money made, and the statements.

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2 years ago

Longer<br> congruent<br> shorter<br> proportional

irinina [24]

Shorter. Because of the way the arches work. the pink dot is the center.

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2 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

RideAnS [48]

Answer:

73 - 2x = 86 - 3x

So 3x - 2x =86 - 73 so x=13

Step-by-step explanation:

Then m<1 = 73-2(13) = 47

M<2=82-3(13)=47

So both angles are having same value

Please Brainlist me

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2 years ago

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

Sliva [168]

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

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

Evaluate the expressions

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

Read more about probability density function at:

brainly.com/question/15318348

#SPJ1

<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

7 0

1 year ago

The researcher has limited resources. He sends 9 emails from a Latino name, and 14 emails from a non-Latino name. For the Latino

deff fn [24]

Answer: 38.41 minutes

Step-by-step explanation:

The standard error for the difference in means is given by :-

$SE.=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2}}$

where , $\sigma_1$ = Standard deviation for sample 1.

$n_1$ = Size of sample 1.

$\sigma_2$ = Standard deviation for sample 2.

$n_2$ = Size of sample 2.

Let the sample of Latino name is first and non -Latino is second.

As per given , we have

$\sigma_1=82$

$n_1=9$

$\sigma_2=101$

$n_2=14$

The standard error for the difference in means will be :

$SE.=\sqrt{\dfrac{(82)^2}{9}+\dfrac{(101)^2}{14}}$

$SE.=\sqrt{\dfrac{6724}{9}+\dfrac{10201}{14}}$

$SE.=\sqrt{747.111111111+728.642857143}$

$SE.=\sqrt{1475.75396825}=38.4155433158\approx38.41$

Hence, the standard error for the difference in means =38.41 minutes

3 0

2 years ago