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ELEN [110]
3 years ago
8

A company purchased a machine for $970,000. the machine has a useful life of 12 years and a residual value of $4,500. it is esti

mated that the machine could produce 1,000,000 units over its useful life. in the first year, 200,000 units were produced. in the second year, production increased to 300,000 units. using the units-of-production method, what is the book value of this asset at the end of the second year of operations?
Mathematics
1 answer:
harina [27]3 years ago
8 0
Cost less salvage value = 970,000 - 4500 = 965,500
Capacity of machine = 1,000,000 units.
units consumed at the end of second year = 200,000 + 300,000 = 500,000 units.
Capacity remaining = 1,000,000 - 500,000 = 500,000 units
Book value at end of second year = (500,000/1,000,000)*965,500 + 4500
= $487,250
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If A and B are supplementary angles, then:
A+B=180º


In this case:
A=44º

44º+B=180º
B=180º-44º=136º

Anser: the measure of the other angle is 136º
7 0
3 years ago
Circles multiple choice<br>geometry <br>please help<br>​
Alexxx [7]

Answer:

a) 12 ( I'm pretty sure)

Step-by-step explanation:

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3 years ago
Choose the equation that represents the graph below:
ehidna [41]
Y=x+3 because when x=0 y=3
5 0
4 years ago
Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars
dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = \dfrac{1}{5} \ car / min = 0.2 \ car /min

the rate of the buses = \dfrac{1}{10} \ bus / min = 0.1 \ bus /min

the rate of motorcycle = \dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min

The probability of any event at a given time t can be expressed as:

P(event  \ (x) \  in  \ time \  (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}

∴

(a)

P(2 \ car \  in  \ 20 \  min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}

P(2 \ car \  in  \ 20 \  min) =0.1465

P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}

P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422

P ( 0 \ buses  \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}

P ( 0 \ buses  \ in \ 20 \ min) =  0.1353

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

= 0.3333 \times \dfrac{1}{4}

= 0.0833

∴

P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}

P(zero  exact  change  in  10 minutes) = 0.4347

c)

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

P( 4  \ motorcyles \  in  \ 45  \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}

P( 4  \ motorcyles \  in  \ 45  \ minutes) =0.0469

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

d)

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle  other vehicle arrives within 5 minutes is)

= \dfrac{6}{12} = 0.5

<u>For motorcycle:</u>

= \dfrac{2 }{12}  = \dfrac{1 }{6}

∴

The required probability = 1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!}  \Bigg)

= 1- 0.5134

= 0.4866

6 0
3 years ago
A number y increased by 5 is at least -21
gayaneshka [121]

                - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - -

\diamond\large\blue\textsf{\textbf{\underline{\underline{Given question:-}}}}

             ❖ A number y increased by 5 is at least -21.

\diamond\large\blue\textsf{\textbf{\underline{\underline{Answer and how to solve:-}}}}

          ❖ First, if you come across the word "increased" it means we

    "increase" a number, or add something to that number, like

y increased by 5 \longmapsto \sf{y+5}

Now, we are also given that this expression is at least -21, which means it can't be -21; it can be -21 or it can be greater than -21.

So we have

\longmapsto\sf{y+5\geq -21}}

<em>Solving for y</em>

<em></em>

  •   Subtract 5 on both sides:-

\longmapsto\sf{y\geq -21-5}

\longmapsto\sf{y\geq -26}

So the values of y greater than or equal to -26 will make this inequality true.

<h3>Good luck with your studies.</h3>

 \rule{300}{1}

7 0
2 years ago
Read 2 more answers
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