Josephine purchased a used vehicle that depreciates under a straight-line
2 answers:
Answer:
D. $4000
Step-by-step explanation:
Initial value of the car is given $5000
Salvage value given $500
Useful life of car given is 5 years
Formula
Depreciation for n years is given as:
Plug in 2 for n. This gives,
∴ Depreciation after 2 years =$ 1000
Value of car after 2 years = Initial value - Depreciation after 2 years
Value of car after 2 years =
Therefore, value of car after 2 years = $ 4000
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We have the mean, μ = 20 and the standard deviation, σ = 4
X ≈ N(20, 4²)
We want P(X<15)
Standardising gives
P(X<15) = P(Z<
)
P(X<15) = P(Z< -1.25) ⇒ When the value of the z-score is negative, we will read the value of z<1.25 on the z-table then subtract the answer from 1
P(X<15) = 1 - P(Z<1.25)
P(X<15) = 1 - 0.8944
P(X<15) = 0.1056
Hence, the probability that pizza takes less than 15 minutes to be delivered is 0.1056 = 10.56%
X ≈ N(20, 4²)
We want P(X<15)
Standardising gives
P(X<15) = P(Z<
P(X<15) = P(Z< -1.25) ⇒ When the value of the z-score is negative, we will read the value of z<1.25 on the z-table then subtract the answer from 1
P(X<15) = 1 - P(Z<1.25)
P(X<15) = 1 - 0.8944
P(X<15) = 0.1056
Hence, the probability that pizza takes less than 15 minutes to be delivered is 0.1056 = 10.56%