Answer:
Mean: 40.17 years.
Standard deviation: 10.97 years.
Step-by-step explanation:
The frequency distribution is in the attached image.
We can calculate the mean adding the multiplication of midpoints of each class and frequency, and dividing by the sample size.
The midpoints of a class is calculated as the average of the bounds of the class.
Then, the mean can be written as:

The standard deviation can be calculated as:
![s=\sqrt{\dfrac{1}{N-1}\sum f_i(X_i-E(X))^2}\\\\\\s=\sqrt{\dfrac{1}{59}[3(15-40.17)^2+7(25-40.17)^2+18(35-40.17)^2+20(45-40.17)^2+12(55-40.17)^2]}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7BN-1%7D%5Csum%20f_i%28X_i-E%28X%29%29%5E2%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B59%7D%5B3%2815-40.17%29%5E2%2B7%2825-40.17%29%5E2%2B18%2835-40.17%29%5E2%2B20%2845-40.17%29%5E2%2B12%2855-40.17%29%5E2%5D%7D)

Answer:
Ok so i'm pretty sure its 3125x^5y^5
Step-by-step explanation:
Answer:
see explaination
Step-by-step explanation:
The p-value for Factor A is greater than 0.1
What is your conclusion with respect to Factor A?
Factor A is not significant.
The p-value for Factor B is greater than 0.1
What is your conclusion with respect to Factor B?
Factor B is not significant.
The p-value for the interaction of factors A and B is greater than 0.1
What is your conclusion with respect to the interaction of Factors A and B?
The interaction of Factors A and B is not significant.
What is your recommendation to the amusement park?
Since method is not a significant factor, use either loading and unloading method.
Let
x = loaves of bread
y = batches of muffins
You must make a system of two equations with two unknowns that describe the problem
3.5x + 2.5y = 17 --- (1)
0.75x + 0.75y = 4.5 --- (2)
Resolving we have
x = 6-y (from (2))
replacing in (1)
3.5 (6-y) + 2.5y = 17
21 - 3.5y + 2.5y = 17
y = 21-17 = 4
Then substituting in (2)
x = 6-y = 6-4 = 2
Answer
Helena could bake:
2 loaves of bread
4 batches of muffins
I think these are the sums of perfect cubes.
A = (2x²)³ + (3)³
B = (x³)³ + (1)³
D = (x²)³ + (x)³
E = (3x³)³ + (x^4)³