Asupermarket chain sells oranges, apples, peaches, and bananas in three stores located throughout a large metropolitan area. the average number of pounds sold per day in each store is summarized in matrix m. “in season” and “out of season” prices, per pound, of each fruit are given in matrix n. what is the total, for the three stores, of “in season” daily revenue for the four fruits? the “out of season” peach sales represent what percentage of the daily total “out of season” revenues for store 3?

1877

64%

Step-by-step explanation:

M =  O   A   P   B

         35 50 30 45     Store 1

         25 30 40 40     Store 2

         70 50 60 55     Store 3

N =  ʺIn seasonʺ ʺOut of seasonʺ

              $3.30             $2.00

              $1.50            $5.70

              $4.20            $4.55

              $5.05            $4.00

The total for the three stores, of “in season” daily revenue for the four fruits

35* 3.30 + 50* 1.50 + 30* 4.20 + 45* 5.05 (store 1)

= 543.75

25* 3.30 + 30* 1.50 + 40* 4.20 + 40* 5.05 (store 2)

= 497.5

70* 3.30 + 50* 1.50 + 60* 4.20 + 55* 5.05 (store 3)

= 835.75

543.75 + 497.5 + 835.75 = 1877

Percentage of the “out of season” peach sales represent the daily total “out of season” revenues for store 3

Total “out of season” revenues for store 3

= 70* 2.00 + 50* 5.70 + 60* 4.55 + 55* 4.00

= 918

Total peaches sale in "out of season"

= 30(4.55) + 40(4.55) + 60(4.55)

= 591.5

Percentage = 591.5/918 * 100%

                    = 64%

D

Step-by-step explanation:

The red number cube has a 1/6 probability of showing a 3 and the blue number cube also has a 1/6 probability of showing a 6.

Just multiply 1/6 times 1/6 to get the probability of both outcomes.