It’s would be 24 photos per day
The z-score of the information given is:
z=(x-μ)/σ
where
μ-mean
σ- standard deviation
from the information given:
x=93.02
μ=89
σ=√625=25
hence
z=(93.02-89)/25
z=0.1608
thus
P(X≥x)=1-0.5636=0.4364=43.64%
Answer:
Total cost is $205328
Step-by-step explanation:
Given data:
cost C(q) = 0.1 q^3 - 0.5 q^2 + 500 q + 200
current units q = 4 ( 4000 units )
The current level of production is 4000 ( 4 )units, and manufacturer is planning to upgrade this to 4100 ( 4.1 ) units
C'(q) = 3 * 0.1 q^2 - 2 * 0.5 q + 500
C'(q) = 0.3 q2 - q + 500
C(4.1) - C(4) ≈ C’(4) x Δq
≈ C’(4) x 0.1
≈ ( 0.3 * 4^2 - 4 + 500 ) x 0.1
≈ ( 0.3 * 16 - 4 + 500 ) x 0.1
≈ 500.8 x 0.1 ≈ 50.08
total cost = 4100 x 50.08 = $205328
