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3 years ago

Can someone help me in number 17 and 17.A and 17.B plzz

Mathematics

2 answers:

Brut [27]3 years ago

7 0

(2+6)x(6+3)=72in^2
And 6+2=8 /6+3=9

Varvara68 [4.7K]3 years ago

5 0

A) (2+6) x (6+3) = 72 in^2
b) width: 6 + 2 = 8
length: 6 + 3 = 9

Because the green part is a square, so the width is just 6 in plus 2 in.
Now, we also know the height of the blue rectangle, 8 in.
The area of the blue rectangle is 24in^2 --> 24/8 = 3. The width of the blue rectangle is 3in.
3 + 6 is the length of the entire figure

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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

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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

7 0

2 years ago

Both circle A and circle B have a central angle measuring 140°. The ratio of the radius of circle A to the radius of circle B is

beks73 [17]

Let
rA--------> radius of the circle A
rB-------> radius of the circle B
LA------> <span>the length of the intercepted arc for circle A
</span>LB------> the length of the intercepted arc for circle B

we have that
rA/rB=2/3--------> rB/rA=3/2
LA=(3/4)<span>π
</span>
we know that
if <span>Both circle A and circle B have a central angle , the ratio of the radius of circle A to the radius of circle B is equals to the ratio of the length of circle A to the length of circle B
</span>rA/rB=LA/LB--------> LB=LA*rB/rA-----> [(3/4)π*3/2]----> 9/8π

the answer is
the length of the intercepted arc for circle B is 9/8π

3 0

3 years ago

My bed is 3 inches by 2and 2/3 inches by 1/3<br> inch. What is the volume of my bed?

ki77a [65]

Step-by-step explanation:

$volume \: of \: bed \\ \\ = 3 \times 2 \frac{2}{3} \times \frac{1}{3} \\ \\ = 3 \times \frac{8}{3} \times \frac{1}{3} \\ \\ = \frac{8}{3} \\ \\ = 2 \frac{2}{3} \: {inch}^{3}$

7 0

3 years ago

Read 2 more answers

What is the Lcm of 28 and 32

lesantik [10]

Answer:

The lcm of 28 is 4

The lcm for 32 is also 4

Step-by-step explanation:

6 0

3 years ago

Given g(x) = 1.1845 - log x, determine for what value of x that g(x) = 1.

MariettaO [177]

When no base is stated, assume base 10
and $log_a(b)=c$ translates to $a^c=b$

g(x)=1, hmm

1=1.1845-log x
times -1 both sides
-1=log x-1.1845
add 1.1845 to both sides
0.1845=log x
translate and assume base 10
$10^{0.1845}=x$
that's the exact value of x such that g(x)=1
you can evaluate it yourself

6 0

3 years ago