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2 years ago

State domain of -5 0 , -4 2, 0 10 , 3 16

1 answer:

7 0

The domain is another name for the values of x

In the points ( -5 , 0) , ( -4 , 2) , (0 , 10) , and (3 , 16)

-5, -4, 0, and 3 are the domains

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A finance magazine did a survey and found that the average American family spends $1,600 on a summer vacation. If the distributi

Andre45 [30]

Answer:

Step-by-step explanation:

Let X be the normal distribution

Kindly find attached a detailed annotation of the solution to the problem.

in the attached document, we found the value that corresponds to 80th percentile, hence we found the value such that the probability that the variable x is lower than this value is 0.8.

Also, from the calculation, we found that the value which corresponds to the 80th percentile is $1945

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2 years ago

Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

Sindrei [870]

Answer:

a) $P(X=3) = 0.1$

b) $P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

c) $P(X=4) = 0.3$

d) $P(X=0) = 0.2$

e) $E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

f) $E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

Step-by-step explanation:

We have the following distribution

x 0 1 2 3 4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

$P(X=3) = 0.1$

Part b

We want this probability:

$P(X\geq 3) =1-P(X$

And replacing we got:

$P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4$

Part c

For this case we want this probability:

$P(X=4) = 0.3$

Part d

$P(X=0) = 0.2$

Part e

We can find the mean with this formula:

$E(X)= \sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2$

Part f

We can find the second moment with this formula

$E(X^2)= \sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4$

And the variance would be:

$Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4$

And the deviation:

$\sigma =\sqrt{2.4} = 1.549$

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2 years ago

Rita had $9.00. She spent $3.67 on a magazine.

Ket [755]

She has $5.33 left

$9.00
-$3.67
————
$5.33

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2 years ago

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The height of the trapezoid is 2 cm

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30in I know that because I just did this .this is EASY!

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