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3 years ago

What is the true hourly wage of a job that pays 18 per hour and allows 20 hours of paid time off for every 500 hours worked?

1 answer:

7 0

To solve for the true hourly wage simply add the 2 rates together.

20 dollars/500 hours = 2/50 = 0.04 dollars per hour.

The true rate would be $18.000 + $0.04 = $18.04 per hour.

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What is the solution to the equation? 5(n-1/10)=1/2?

yarga [219]

5(n-1/10)=1/2=n=1/5 OR n=0.2

5 0

3 years ago

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Wilma can mow a lawn in 80 mintues.Melissa can mow the same lawn in 120 minutes. How long does it take for both Wilma and Meliss

Luba_88 [7]

There is a formula for that;
Total time = (Worker 1 * Worker 2) / (Worker 1 + Worker 2)
Total time = (80 * 120) / (80 + 120)
Total time = 9,600 / 200
Total time = 48 minutes

4 0

3 years ago

A cereal company is putting 1 of 3 prizes in each box of cereal. The prizes are evenly distributed so the

Fittoniya [83]

Answer:

0.625

Explanation:

The image below was not posted for the but is needed to anser the question.

Further Explanation:

Mohamed carried out 24 trials of a simulation

Therefore, Total frequencies = 24

To get number of frequency greater than 3 we have to use the result in the result in the graph to obtain the estimate. Look at the graph below, when you count the dots that are on number of boxes purchased greater than 3, you will get 15.

Therefore, Number of frequency greater than 3 = 15

P(more than 3 boxes) = 15/24 = 0.625

3 0

3 years ago

What is the Quotient of 1.65 of 10

Digiron [165]

$The\ quotient\ of\ 1.65\ of\ 10=1.65:10=\dfrac{1.65}{10}=\dfrac{1.65\times100}{10\times100}\\\\=\dfrac{165}{1,000}=\huge\boxed{0.165}\\\\or\ other\ method\\\\1.65:10\to move\ the\ point\ one\ place\ to\ the\ left\\1.65:10=\huge\boxed{0.165}\\\\if\ 1.65:100\ then\ move\ the\ point\ two\ places\ to\ the\ left.$

5 0

4 years ago

Read 2 more answers

Find the exact value of the expression.

Tresset [83]

$\sin(a-b)=\sin a \cos b-\cos a \sin b$ . If we let $a=\cos^{-1} \left(\frac{5}{6} \right)$ and $b=\tan^{-1} \left(\frac{1}{2} \right)$ , then the given expression is equal to:

$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$ ,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

$2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$

6 0

2 years ago

Read 2 more answers