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stich3 [128]
3 years ago
13

An office has 80 employees, and 24 of the employees are managers. What percentage of the employees are managers?

Mathematics
2 answers:
Viktor [21]3 years ago
8 0

Solution:

30% of the employees are managers.

Explanation:

We have been given that there are 80 employees, and 24 of the employees are managers.

We have to find the percentage of the employees that are managers.

We can find this percentage with the below formula

\% = \frac{\text{Number of managers}}{\text{Total number of employee}}\times 100

On substituting the known values, we get

\%=\frac{24}{80}\times 100\\\%=\frac{3}{10}\times 100\\\%=3\times 10\\\%=30\%

Therefore, 30% of the employees are managers.

hichkok12 [17]3 years ago
3 0
You would have to do the problem doing:

X/100 = 24/80

And cross multiply 100 and 24. Then you would divide by the 80 and that leads to the answer of 30%.
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Katyanochek1 [597]

Answer:

  1. b/a
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  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

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1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

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3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
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