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34kurt
3 years ago
13

Given the pay rate and hours worked, determine the gross earnings, Federal taxes (assuming 20% of gross earnings), state taxes (

assuming 6% of gross earnings), social security deduction (assuming 6.45% of gross earnings), total deductions, and net pay.
Mathematics
1 answer:
qaws [65]3 years ago
8 0

Answer: Total deductions= 0.3245ab and Net pay= 0.6755ab


Step-by-step explanation:

Suppose, the pay rate= a dollar per hour

And hours worked= b hours

Gross earning= the pay rate x hours worked

Gross earning= ab

Deductions are:

Federal taxes= 20% of gross earnings= 20/100 x ab= 0.2ab dollars

State taxes= 6% of gross earning= 6/100 x ab= 0.06ab dollars

Social security deduction= 6.45% of gross earning= 6.45/100 x ab= 0.0645ab

So,

The total deduction= federal taxes + state taxes + social security deduction

The total deduction= 0.2ab + 0.06ab + 0.0645ab= 0.3245ab dollars

And  

Net pay= gross earning – total deduction

Net pay= ab – 0.3245ab= 0.6755ab dollars


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a car travels 60km in 36minutes.At the same average speed, how far will it travel in 1 hour 12 minutes​
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5 0
3 years ago
A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
4 years ago
Select the correct answer
NemiM [27]

let f(r)=sin^3r/cos^3r

so f(-r)= (sin(-r)/cos(-r))^3

=(-sinr/cosr)^3

=-f(r)

so it is odd function

4 0
3 years ago
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