2(5*10)+2(10*4)+2(5*4)=
2(50)+2(40)+2(20)=100+80+40=
220 cm^2
I found the answer by finding the areas of the 3 different sides, multiplying each area by 2 since each side has another side that is equal on a rectangular prism, then adding all of them together.
Another name for a relation is a function.
As a ratio, 5:86 represents students to teachers
We can set up the equation as 5/86=x/9460
Multiply 9460 to both sides
5x9460/86=x
47300/86=x
x=550
There are 550 professors
What is the solution set of x2 + y2 = 26 and x − y = 6? A. {(5, -1), (-5, 1)} B. {(1, 5), (5, 1)} C. {(-1, 5), (1, -5)} D. {(5,
Rus_ich [418]
He two equations given are
x^2 + y^2 = 26
And
x - y = 6
x = y +6
Putting the value of x from the second equation to the first equation, we get
x^2 + y^2 = 26
(y + 6) ^2 + y^2 = 26
y^2 + 12y + 36 + y^2 = 26
2y^2 + 12y + 36 - 26 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
y^2 + y + 5y + 5 = 0
y(y + 1) + 5 ( y + 1) = 0
(y + 1)(y + 5) = 0
Then
y + 1 = 0
y = -1
so x - y = 6
x + 1 = 6
x = 5
Or
y + 5 = 0
y = - 5
Again x = 1
So the solutions would be (-1, 5), (1 , -5). The correct option is option "C".
Answer:
And using the normal standard table or excel we find the probability:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the avergae number of weeks an individual is unemployed of a population, and for this case we know the distribution for X is given by:
Where 
and 
Since the distribution for X is normal then, the distribution for the sample mean 
is given by:
We select a sample of n =50 people. And we want to find the following probability
And using the normal standard table or excel we find the probability:
