Help help help please and thank you

n a survey of 331 customers, 66 say that service is poor. You select two customers without replacement to get more information o

yaroslaw [1]

Answer:

3.93% probability that both say service is poor

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The customers are chosen without replacement, and the order in which they are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

What is the probability that both say service is poor?

Desired outcomes:

Two saying it is poor, from a set of 66. So

$D = C_{66,2} = \frac{66!}{2!(66-2)!} = 2145$

Total outcomes:

Two customers from a set of 331. So

$T = C_{331,2} = \frac{331!}{2!(331-2)!} = 54615$

Probability:

$p = \frac{D}{T} = \frac{2145}{54615} = 0.0393$

3.93% probability that both say service is poor

4 0

2 years ago

Add 3 and 7. then multiply by 3.

Reil [10]

Answer:

30

Step-by-step explanation:

First, add 3 and 7

3 + 7 =10

The multiply by 3

10*3= 30

Hope this helps!

8 0

3 years ago

Read 2 more answers

Please need help asap!!

vagabundo [1.1K]

$a+b+c=180, \text{ }c+e=180 \implies \\ a+b+c=c+e \implies \\ a+b=e\\$

This result is actually true for any exterior angle. The exterior angle of a triangle is equal to the sum of the two remote angles, and above is a short proof of it.

6 0

3 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

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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

Using the rules for special cases, what is the result of?

Arlecino [84]

Answer:

C) 4x^2 -8x +4[/tex]

Step-by-step explanation:

(-2x+2)^2

we apply (a+b)^2 formula

$(a+b)^2 = a^2 + 2ab +b^2$

a= -2x and b= 2

Replace 'a' with -2x and 'b' with 2

$(-2x+2)^2 = (-2x)^2 + 2(-2x)(2) +2^2$

(-2x)^2 is 4x^2

$(-2x+2)^2 = 4x^2 -8x +4$

option C is the answer

8 0

3 years ago

Read 2 more answers