What is the question?
I'm assuming it is to find the length and width.
+_= plus or minus
(X+36)
____________
| |
(X) | |
|____________|
X^2+36X-2040<0
X<-36+_(36^2-4*-2040)^(1/2)
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2
X<-18+_2((591)^(1/2))
This is probably not what you wanted, sorry
Answer: y+0
As y=0 represent x-axis whose slope is 0 and intersects y-axis at the origin.
Step-by-step explanation:
Answer:
0.108
Step-by-step explanation:
Using the poisson probability process :
Where :
P(x =x) = (e^-λ * λ^x) ÷ x!
Given that :
Each batch of bread = 3 loaves
Each loaf = 15 slices
Total slice per batch = 15 * 3 = 45 slices
Number of raising added = 100
Average number of raisin per slice, λ = 100/45 = 20/9
Hence,
Probability that a randomly chosen slice has no raising :
P(x = 0) = (e^-λ * λ^x) ÷ x!
P(x = 0) = (e^-(100/45) * (100/45)^0) ÷ 0!
P(x = 0) = (0.1083680 * 1) / 1
P(x = 0) = 0.108
Answer:
18, 21, 36
Step-by-step explanation:
Let L represent the least number. Then the greatest is 2L and the middle number is (L+3). Their sum is ...
L +(2L) +(L+3) = 75
4L = 72 . . . . . . . . . subtract 2, collect terms
L = 18 . . . . . . . . . . . divide by 4
L+3 = 21
2L = 36
The numbers are 18, 21, and 36.
Answer:
The standard deviation of the number of rushing yards for the running backs that season is 350.
Step-by-step explanation:
Consider the provided information.
The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards.
Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.
Use the formula: 
Here z is 2.42 and μ is 790, substitute the respective values as shown.
Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.
