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ladessa [460]
2 years ago
7

Suppose two random variables, X and Y are independent, which statement is false? a) P(X∣Y)=P(X) b) P(X∪Y)=P(X)+P(Y) c) P(X∩Y)=P(

X)⋅P(Y) d) Cov(X,Y)=0
Mathematics
1 answer:
Svetlanka [38]2 years ago
5 0

B is false, since by the inclusion/exclusion principle,

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)

By independence, we have P(X\cap Y)=P(X)P(Y), which is zero if either of P(X) or P(Y) is 0, which isn't guaranteed.

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Z+x3:use x = 1 and z = 19
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3 years ago
Select the correct answer.
Marta_Voda [28]
Remember x is x and f(x) is same thing as y.

f(5) = 4

It is same thing as saying, when x = 5, y aka f(x) should be 4. Aka the point (5,4)

Let’s look at D, if x =5 then f(x)=5? No, it should equal 4. So this is wrong. It is also not linear. So NOT D

Lets look at C, if x =5 then
1/5(5)+3 =
5/5 + 3 =
1+3 =
4

So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, but it is not linear. So NOT C

Lets look at B, if x=5 then,
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So, f(x) aka y = 4. Is this correct? Yes when x =5, y indeed should equal 4, AND it is linear SO THE ANSWER IS B

Check A too, it gets y=14 so thats wrong as it is not 4.


5 0
1 year ago
You are out bottle caps, and decide to use Popsicle sticks instead. You measure them, and they are 15.2 cm tall. How many Popsic
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3 years ago
The dimensions of a 4-in. square are multiplied by 3. How is the area affected? {Show work!}
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If dimensions are multiplied by 3 then area is multiplied by 3^2
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