A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obtained only by surgery. Suppose that a CAT scan of a human liver shows cross-sections spaced 1.5 cm apart. The liver is 15 cm long and the cross-sectional areas, in square centimeters, are 0, 19, 58, 78, 94, 105, 117, 129, 62, 40, and 0. Use the Midpoint Rule with n = 5 to estimate the volume V of the liver. V = cm3

Step-by-step explanation:

Here we have

Space in cm          Area in square centimeters  -

                                    19                   58                78                   94               105                  117              129                  62                40                     0

As we can see, and we need to divide it into 5 subintervals, where

                                         

is the length of each interval. Therefore, the subintervals are

                                 

The midpoint of the first interval is calculated as

                                       

Similarly, we obtain that the midpoints of other subintervals are

                             

By the Midpoint rule, we have

                     

From the given data above, we have

         

Therefore,

                     

Probability that it does not rain during the entire festival = 0.19584

Step-by-step explanation:

We are given that you and a group of friends are going to a five-day outdoor music festival during spring break.

Also, Probability that there may be rain on first day, P(A) = 0.15

Probability that there may be rain on second day, P(B) = 0.10

Probability that there may be rain on third day, P(C) = 0.20

Probability that there may be rain on fourth day, P(D) = 0.20

Probability that there may be rain on fifth day, P(E) = 0.60

It is also provided that these probabilities are independent of whether it rained on the previous day or not.

Now, probability that it does not rain during the entire festival = Probability that there may not be rain on all five days

= (1 - P(A)) (1 - P(B)) (1 - P(C)) (1 - P(D)) (1 - P(E))

= (1 - 0.15) (1 - 0.10) (1 - 0.20) (1 - 0.20) (1 - 0.60)

= 0.85 0.90 0.80 0.80 0.40 = 0.19584