Lindsay Electronics, a small manufacturer of electronic research equipment, has approximately 6 comma 500 items in its inventory and has hired Joan Blasco-Paul to manage its inventory. Joan has determined that 11% of the items in inventory are A items, 33% are B items, and 56% are C items. She would like to set up a system in which all A items are counted monthly (every 20 working days), all B items are counted quarterly (every 62 working days), and all C items are counted semiannually (every 121 working days). How many items need to be counted each day? The total number of items that need to be counted each day is nothing items (round your response to the nearest whole number).

The total number of items to be counted each day ≈ 100

Step-by-step explanation:

Lindsay Electronics has 6,500 items in its inventory.

11% of the items in inventory are A items

33% of the items in inventory are B items

56% of the items in inventory are C items

A items are counted every 20 working days

B items are counted every 62 working days

C items are counted every 121 working days

How many items need to be counted each day?

First we will find the number of items of type A, B and C

Number of A items = 11% of 6,500 = 0.11*6500 = 715

Number of B items = 33% of 6,500 = 0.33*6500 = 2145

Number of C items = 56% of 6,500 = 0.56*6500 = 3640

The number of A items to be counted each day is

A items = 715/20

The number of B items to be counted each day is

B items = 2145/62

The number of C items to be counted each day is

C items = 3640/121

The total number of items to be counted each day is

Total items = 715/20 + 2145/62 + 3640/121

Total items = 100.42

Rounding the answer to the nearest whole number yields,

Total items ≈ 100

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation:

lolololkzaxio2ier286udeg