Answer:
a)= 2
b) 6.324
c) P= 0.1217
Step-by-step explanation:
a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)
b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by
σ_X`1- X`2 = √σ²/n1 +σ²/n2
Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2
so
σ_X`1- X`2 =√20 +20 = 6.324
if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.
Where as Z is normally distributed with mean zero and unit variance.
If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then
Z= 0-2/ 6.342= -0.31625
P(-0.31625<z<0)= 0.1217
The probability would be 0.1217
Answer:
B. -76-7C
Step-by-step explanation:
-6-7(c+10)
Distribute the 7
-6-7c-70
Combine like terms
-7c-76
The number = y
2y + 1/3 = 2/3 - 3y
(add 3y to both sides)
5y + 1/3 = 2/3
(subtract 1/3 from both sides to isolate y)
5y = 1/3
(divide by 5)
y = 1/15
The answer is A, -840.
Hope this helps!
Answer: About 800 thousand
The more accurate value is 807,528 but this is also an approximation.
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Work Shown:
That's the rough population of wild pigs (in millions) for the year 2000.
Multiply by 
to get it in terms of units instead.
There were roughly 800 thousand wild pigs in the year 2000.
