6p+10p+15-9
16p+6 will be your final answer
With the b2+b- 12 your gonna want to do this
b2+b-12
b2+4b-3b-12
which is the sum product
after doing that you wanna common the factors from the two pairs. Which is b2+4b-3b-12 then you do you parentheses around b(b+4)-3(b+4) just like that after you do that then you rewrite it in factored form b(b+4)-3(b+4) you then want to rewrite it to this (b-3)(b+4) after that your solution will be (b-3)(b+4) for the first one.
Answer:
The expected total amount of time the operator will spend on the calls each day is of 210 minutes.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
n-values of normal variable:
Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is
and the standard deviation is 
Calls to a customer service center last on average 2.8 minutes.
This means that 
75 calls each day.
This means that 
What is the expected total amount of time in minutes the operator will spend on the calls each day
This is M, so:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.
The money altogether is 8.00$
The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.
If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.
The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.
The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ = coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.
This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force to overcome the centripetal force.