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2 years ago

Crane Company provided the following information from its accounting records for 2019.

Mathematics

1 answer:

Ipatiy [6.2K]2 years ago

8 0

Answer:

$10.00 per hour

Step-by-step explanation:

Overhead application rate which is also known as overhead absorption rate on the basis on labor hours is the total budgeted overhead for 2019 which is $900,000 divided by the expected production of 90,000 labor hours for the year.

overhead application rate=$900,000/90,000=$10 per hour

This implies that for every one hour worked overhead cost of $10 would be added to the other costs incurred.

The correct option then is the third option of $10.00 per hour

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Harman [31]

I hope this helps you

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2 years ago

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the sum of two numbers is 67 if three times the smaller number is subtracted by the larger the result is 7, find two numbers

just olya [345]

Um I think it's 3x7x4=81 or like this 40+20+7=67.

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3 years ago

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What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

Solution:

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$

As for infinitely many solutions $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}$

Hence b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

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2 years ago

Bryan sells cars. He makes a salary of $500 a week, plus $150 for every car he sells. This week, he wants to make at least $2000

Kobotan [32]

If Bryan wants to make $2000 dollars then he needs to solve this equation:

2000=500+150x
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1500=150x divide by 150 on both sides

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natima [27]

C would be my guess here

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