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vladimir1956 [14]
3 years ago
6

A compact minivan costs $16,000 with a residual value of $1,000. It has an estimated useful life of five years. If the minivan w

as bought on July 3, what would be the book value at the end of Year 1 using straight-line rate? A. $14,500 B. $16,000 C. $1,500 D. $12,500
Mathematics
1 answer:
Andreyy893 years ago
7 0

Answer:

  A.  $14,500

Step-by-step explanation:

The van depreciates ($16000 -1000 = $15000 in 5 years, so $3000 per year. It will be assumed to depreciate half that amount in half a year, so will be worth $1500 less than $16000 at the end of the first calendar year. The book value will be $14,500.

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Could someone please explain to me how to solve this problem? Thanks! :)
BartSMP [9]

Answer:

First of all since they are dividing subtract the powers of the base from each other .

Make sure the bases are the same.

((6^7-6)(3^3-4))³

((6)(3^-1))³

(6³)(3^-3)

216 × 1/27

= 8

That's option d.

Hope this helps.

6 0
3 years ago
May somebody help me with this please?
beks73 [17]
First you do 4500 × 0.04
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2 years ago
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Solve the following for the specified variable<br><br> Z=s/2(P+p); for P
lapo4ka [179]

Answer:

\boxed{\sf P = \frac{2Z}{s} - p}

Step-by-step explanation:

\sf Solve \ for \ P: \\ \sf \implies  Z = \frac{s}{2} (P + p) \\ \\ \sf Z = \frac{s}{2} (P + p) \ is \ equivalent \ to \ \frac{s}{2} (P + p) = Z: \\ \sf \implies \frac{s}{2} (P + p) = Z \\ \\  \sf Divide \ both \ sides \ by \ \frac{s}{2} : \\ \sf \implies P + p = \frac{2Z}{s} \\ \\ \sf Substrate \ p \ from \ both \ sides: \\ \sf \implies P = \frac{2Z}{s} - p

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2 years ago
Determine whether the data described in the tables is best modeled by a linear function or an exponential function.
vova2212 [387]

Answer: Tigers: Linear Function

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2 years ago
Zucchini weights are approximately normally distributed with mean 0.8 pound and standard deviation 0.25 pound. Which of the foll
Charra [1.4K]

Answer:

0.81859

Step-by-step explanation:

We solve using z score method

The formula for calculating a z-score is is z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Mean 0.8 pound

Standard deviation 0.25 pound

For x = 0.55 pound

z = 0.55 - 0.8/0.25

= -1

Probability -value from Z-Table:

P(x = 0.55) = 0.15866

For x = 1.3 pounds

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Probability value from Z-Table:

P(x = 1.3) = 0.97725

The probability that a randomly selected zucchini will weigh between 0.55 pound and 1.3 pounds is

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0.97725 - 0.15866

0.81859.

6 0
3 years ago
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