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3 years ago

Help!with this question ASAP

Mathematics

1 answer:

Nookie1986 [14]3 years ago

6 0

Answer:

i belive it is B

Step-by-step explanation:

if they only want the jars from the second month you'd need to subtract from the first one

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A twelve-sided die with sides labeled 1 through 12 will be rolled once. Each number is equally likely to be rolled.

kykrilka [37]

Answer:

1/6

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3 years ago

Ben is 6 feet tall. At a certain time of day, Ben's shadow appears on the ground to be 6 feet long. What is the angle of elevati

nevsk [136]

Step-by-step explanation:

let us consider the elevation of the sun is x

then

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3 years ago

The circumference of a circle is 36pi cm what is the radius

alexdok [17]

Answer:

The radius is 18 cm

Step-by-step explanation:

The formula for circumference with radius is

C = 2*pi*r

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36 pi = 2 * pi *r

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4 years ago

Math question down below

julsineya [31]

This data set has no outliers, because all of the numbers are close together.

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3 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago