Answer:
The height of the given rectangular prism is 5cm.
Step-by-step explanation:
Given that the volume of the rectangular prism is 
And length is
and width is 
That is V=5cm^3[/tex] ,
and 
We have the volume of the rectangular prism V=lwh cubic centimeter

Substituting the values in above formula




Rewritting as below

Therefore the height is 5cm
Answer:
It is not normally distributed as it has it main concentration in only one side.
Step-by-step explanation:
So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).
So, let us begin the groupings into their different classes, shall we?
Data given:
0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.
(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.
(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31
(3). 0.4 - 0.6 : there are 2 values that falls into this category.
(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.
Class interval frequency.
0.00 - 0.20. 15.
0.20 - 0.40. 2.
0.4 - 0.6. 2.
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
Population: six sugar maples
Sample: sap
In decimal form that would be 36.052