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Julli [10]
3 years ago
15

You are still debating about the two different computers. If you are paid $10.00/hour and your deductions are FICA (7.65%), fede

ral tax withholding (12%), and state tax withholding (8%), how many hours do you have to work to pay for the new computer?
Include step by step
Mathematics
1 answer:
sattari [20]3 years ago
8 0

Answer:

89 hours

Step-by-step explanation:

First solve all the deductions.

For each multiply the $10, which you are paid, and the percentage.

For FICA: $10 x 0.0765 = $0.77 per hr

For Fed tax: $10 x 0.12 = $1.2 per hr

For State tax: $10  x 0.08 = $0.8 per hr

Now subtract the remaining pay:

10 – 0.77 – 1.2 – 0.8 = $7.23

Time required = $640 / $7.23 per hr

Time required = 89 hrs

Final answer:

89 hours

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36 * 2/3 = 24 were blueberry.

36 - 24 = 12 were not blueberry. 
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Check attachment.

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3 years ago
The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0
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Answer:

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The "Commutative Laws" say:  when we add we can swap numbers over and still get the same answer.

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6 0
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