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Ahat [919]
3 years ago
10

It has been reported that men are more likely than women to participate in online auctions. In a recent​ survey, 65% of responde

nts reported that they had participated in an online auction. In this same​ survey, 45% of respondents were men and​ 38% were men who had participated in online auctions. What is the probability that a respondent had participated in an online auction given that he is​ male?
Mathematics
1 answer:
vivado [14]3 years ago
8 0

Answer:

0.84 or 84%

Step-by-step explanation:

Let 'n' be number of people who took the survey, the number of men who answered the survey is 0.45n. The number of men who answered the survey and have participated in online auctions is 0.38n. Therefore, choosing from the population of men who took the survey (0.45n) the probability that a respondent had participated in an online auction is:

P =\frac{0.38n}{0.45n}\\P=0.84

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Which equation could represent the graph shown below
kaheart [24]

Answer:

f(x) = √(x+1) -2

Step-by-step explanation:

Since you're working with transformed functions, you know that replacing x with x-a in f(x) will translate the graph "a" units to the right. You also know that adding "b" units to the function value will translate the graph "b" units upward.

Here, the graph of f(x) = √x has been translated 1 unit to the left (a=-1) and 2 units down (b=-2). So, the transformed function is ...

f(x) = √(x-(-1)) + (-2)

f(x) = √(x+1) -2 . . . . . . simplify

6 0
4 years ago
What is the sample space for the event of flipping a coin and then spinning a four-part spinner, numbered 1, 2, 3, 4?
Anastasy [175]

Answer:it is 2 since it include3s all possible answers for both heads & tails.

Step-by-step explanation:

you can get a H1-4 & a T1-4 since you have 4 parts one the spinner & 2 things on a coin (Heads & tails) you can <u>just times 4*2 & know that you have 8 possible answers.</u>

8 0
3 years ago
A random variable X with a probability density function () = {^-x &gt; 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
2 years ago
17. Find the reflection of the point A(6.-1) in the x-axis.<br>​
castortr0y [4]

Answer:

Your answer will be (6,1)

Step-by-step explanation:

I know this because a reflection over the x-axis is (x,y) -> (x,-y) If you apply this to (6,-1) you get (6,1).

5 0
3 years ago
F(1)=3f(1)=3 and f(n)=2f(n-1)+3 <br><br> f(n)=2f(n−1)+3 <br><br> then find the value of f(3)
gayaneshka [121]

Answer:

f(3)=21

Step-by-step explanation:

f(1)=3

f(n)=2f(n-1)+3

n=2

f(2)=2f(2-1)+3

f(2)=2f(1)+3

f(2)=2*3+3

f(2)=6+3

f(2)=9

n=3

f(3)=2f(3-1)+3

f(3)=2f(2)+3

f(3)=2*9+3

f(3)=18+3

f(3)=21

7 0
3 years ago
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