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Ksivusya [100]
3 years ago
14

How could you describe the correlation in the data ? Explain your reasoning

Mathematics
1 answer:
ddd [48]3 years ago
5 0

Answer:

We make scatterplots to see relationships between variables. Scatterplots are really good for helping us see if two variables have positive or negative association (or no association at all).Step-by-step explanation:

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the radius of the base of a right cylinder is 12 cm and the height and the height of the cylinder is 18 cm. in terms of pi what
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Answer:

Step-by-step explanation:r=12cm,h=18cm,v=?

Vol. Of a cylinder=Πr²h

V= Π×12²×18

V= Π×144×18

V=2592Πcm³

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Y varies directly with x,and y=7 when x=2. What is the value of y when x=5?
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Y=10 bc y-x=5 then 5+5= 10
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A salesperson earns a commission of $308 for selling $ 2200 in merchandise. Find the commission rate.
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Step-by-step explanation

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A snail can crawl 2/5 of a meter in a minute. How many minutes will it take the snail to crawl 6 meters? Enter your answer in th
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Read 2 more answers
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
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