Answer:
A)-11
-11 is less than -10 so it cant be great than -10
Let x is Jiri's age
2x is Pamela's age
x + 2x = 33
3x =33
x = 33/3
x =11
Answer: Jiri's age is 11
F(x)=6x^4-10x^3+40x-50, plug 2 in for x
f(2)=6(2)^4-10(3)^3+40(2)-50
f(2)=12^4-30^3+80-50
f(2)=20735-27,000+80-50
f(2)=-6,235
Answer:
0.1507 or 15.07%.
Step-by-step explanation:
We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.
First of all, we will find z-scores for data points using z-score formula.
, where,
z = z-score,
x = Sample score,
= Mean,
= Standard deviation.



Let us find z-score of data point 22.03.



Using probability formula
, we will get:

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.
The answer is the 2nd choice, number of sales.
The function described f(x) = B + Cx