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romanna [79]
4 years ago
8

A company has a monthly budget of x dollars. Every month, $20,700 of the monthly budget is spent on salaries. One-fifth of the r

emaining budget is spent on office supplies. Which function can be used to calculate the amount, in dollars, spent on office supplies?
Mathematics
2 answers:
Nastasia [14]4 years ago
6 0

Answer:

OS=\frac{x-20,700}{5}

Step-by-step explanation:

<u>Functions to model reality  </u>

There are situations in real life where some variables depend on other(s) and that dependency can be determined as a practical rule or from a well-known theory. The expression who relates those variables are commonly called functions and they become a very useful tool to analyze, explain and predict the behavior of such events.

We know the monthly budget of a company is x dollars. $20,700 out of it is spent on salaries. After deducting salaries, the remaining budget is

x-20,700

One-fifth of it is spent on office supplies, so the required function is

OS=\frac{1}{5}(x-20,700)

OS=\frac{x-20,700}{5}

12345 [234]4 years ago
4 0

Answer: <em>Amount spent on </em><em>office supplies</em><em> can be calculated as this: </em><em> $(1/5 × (x — 20,700)</em><em>.</em>

Step-by-step explanation:

Step 1:

Let Monthly Budget of the Company, M = $x

Amount spent on salaries, S = $20,700

Step 2:

To get  balance of the budget, R. Subtract S from M

Therefore,

Remaining balance of the budget, R = $(x — 20,700)

Step 3:

Amount spent on office supplies O is given as 1/5 of  balance of the budget. I.e, O = 1/5 × R

                                         

Therefore, O = $(1/5 × (x — 20,700).

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Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place
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Answer:

The Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for \int\limits^5_4 {x^2+4} \, dx with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)

where \Delta{x}=\frac{b-a}{n}

We know that a = 4, b = 5, n = 4.

Therefore, \Delta{x}=\frac{5-4}{4}=\frac{1}{4}

Divide the interval [4, 5] into n = 4 sub-intervals of length \Delta{x}=\frac{1}{4}

\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]

Now, we just evaluate the function at the midpoints:

f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625

f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625

f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625

f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625

Finally, use the Midpoint Sum formula

\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125

This is the sketch of the function and the approximating rectangles.

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