Answer:
5(P - 1.00) + 3(P - 0.75)
Step-by-step explanation:
Cost of a package in store A = $P
Discount offered at store A = $1.00
Price paid per package at store A = Cost of a package in store A - Discount offered at store A
= P - 1.00
Cost of a package in store B = $P
Discount offered at store B = $0.75
Price paid per package at store B = Cost of a package in store B - Discount offered at store B
= P - 0.75
Total cost of 5 packages at store A and 3 packages at store B = 5(P - 1.00) + 3(P - 0.75)
Answer:
Step-by-step explanation:
There are C(5,2) = 10 ways to choose 2 contraband shipments from the 5. There are C(11, 1) = 11 ways to choose a non-contraband shipment from the 11 that are not contraband. Hence there are 10*11 = 110 ways to choose 3 shipments that have 2 contraband shipments among them.
There are C(16,3) = 560 ways to choose 3 shipments from 16. The probability that 2 of those 3 will contain contraband is
110/560 = 11/56 ≈ 19.6%
_____
C(n,k) = n!/(k!(n-k)!)
Answer:
17%
Step-by-step explanation:
Divide $21.25 by $125
Hey There!
If i got this down right,
An = A1 + (n - 1)d
Sn = A1 + A2 + A3 + ... + An
A1 is given by
Sn = n (A1 + An) / 2
For Example
An = A1 + (n - 1)d
= 6 + 3 (n - 1)
= 3 n + 3
n = 50
Hope This Helps!!!
