Answer:
0.5
Step-by-step explanation:
This is an example of mutually exclusive event in which two or more events cannot occur at the same time or in a single performance.
Therefore, the probability the number of cars entering the roadway is between 4 and 6 cars per minute (inclusive) implies that it is either 4, 5 or 6 cars enter in a minute. Since we have an average of 6 cars per minute, the probability of each of the event occurring is 1/6. This can therefore be calculated as follows:
P(4 or 5 or 6) = (1/6) + (1/6) + (1/6)
= 3/6
= 1/2
= 0.5
Therefore, the probability the number of cars entering the roadway is between 4 and 6 cars per minute (inclusive) is 0.5.
Answer:
3,600
Step-by-step explanation:
(22 - 2) x 180
20 x 180
3,600
Section A) The AROC for Part A is 4.
h(x) = 4 * 1. 4 is your answer.
Section B) The AROC for Part B is 1.
h(x) = 4 * 1. 4 is your answer.
Part A: (Above)
Part B: There is no greater average change, both X's AROC is 4.
A.R.O.C: (Average Rate of Change)
I think this is right. Not 100% though
Answer:
b
Step-by-step explanation:
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
