The sales of Universal Instruments in the first t years of its operation are approximated by the function

S(t) = t

0.96t2 + 25

where S(t) is measured in millions of dollars. What were Universal's average yearly sales over its first 5 years of operation?

To find Universal Instruments' average yearly sales over its first 5 years of operation, we need to calculate the average rate of change of the sales function over that time period.

The sales function is given by:

S(t) = 0.96t^2 + 25

We want to find the average of S(t) over the first 5 years, so we need to evaluate it at the endpoints and divide by the length of the interval.

Let's evaluate the function at the endpoints:

S(0) = 0.96(0)^2 + 25 = 25
S(5) = 0.96(5)^2 + 25 = 0.96(25) + 25 = 24 + 25 = 49

Now we can calculate the average yearly sales:

Average sales = (S(5) - S(0)) / 5

= (49 - 25) / 5

= 24 / 5

= 4.8 million dollars

Therefore, Universal Instruments' average yearly sales over its first 5 years of operation is approximately 4.8 million dollars.

To find the average yearly sales over the first 5 years of operation, we need to calculate the total sales over the 5 years and divide it by 5.

Let's calculate the total sales over the first 5 years by summing up the individual sales for each year:

S(1) = 0.96(1)^2 + 25 = 0.96 + 25 = 25.96 million dollars
S(2) = 0.96(2)^2 + 25 = 0.96(4) + 25 = 28.84 million dollars
S(3) = 0.96(3)^2 + 25 = 0.96(9) + 25 = 33.44 million dollars
S(4) = 0.96(4)^2 + 25 = 0.96(16) + 25 = 40.84 million dollars
S(5) = 0.96(5)^2 + 25 = 0.96(25) + 25 = 47 million dollars

Now, we can calculate the average yearly sales by dividing the total sales by 5:

Average yearly sales = (25.96 + 28.84 + 33.44 + 40.84 + 47) / 5 = 175.08 / 5 = 35.02 million dollars

Therefore, Universal Instruments' average yearly sales over its first 5 years of operation were approximately $35.02 million.

the average value of f(t) over the interval [a,b] is

∫[a,b] f(t) dt
------------------
b-a

So, for this problem, that would be

(∫[0,5] .96t^2+25 dt)/(5-0)
= (.32t^3 + 25t [0,5])/5
= 165/5
= 33