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Elanso [62]
3 years ago
6

This year, 1,272 students enrolled in night courses a a local college. Last year only 1,200 students enrolled. What was the perc

ent increase in the enrollment? Which operation will you use in your first step?
Mathematics
1 answer:
Ahat [919]3 years ago
4 0

The percent increase in enrollment is 6 %

The operation used in first step is finding the difference between final value and initial value

<h3><u>Solution:</u></h3>

Given that This year, 1,272 students enrolled in night courses a a local college

Last year only 1,200 students enrolled.

To find: percent increase in the enrollment

The percent increase between two values is the difference between a final value and an initial value, expressed as a percentage of the initial value.

<em><u>The percent increase is given as:</u></em>

\text { percentage increase }=\frac{\text { final value - initial value}}{\text { initial value }} \times 100

Here initial value (last year) = 1200 and final value(this year) = 1272

Substituting the values in above formula,

\begin{aligned}&\text { percentage increase }=\frac{1272-1200}{1200} \times 100\\\\&\text { percentage increase }=\frac{72}{1200} \times 100=6\end{aligned}

Thus percent increase is 6 %

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Hermes says that the opposite of 9 is 9 since
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Hope this helps! Comment below for more questions.
3 0
3 years ago
Factored 6x2 – 31x - 30 ​
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3 years ago
(4/2)4 + (1 + 3)?<br><br> I really need help!
Katyanochek1 [597]

\huge\text{Hey there!}

\large\text{Just do PEMDAS}

\large\text{Parentheses}

\large\text{Exponents}

\large\text{Multiplication}

\large\text{Division}

\large\text{Addition}

\large\text{Subtraction}

\mathsf{(\dfrac{4}{2})(4) + (1+3)}

\mathsf{\dfrac{4}{2} = 2}

\mathsf{2(4) + (1 + 3)}

\mathsf{2(4) = 8}

\mathsf{8 + (1 + 3)}

\mathsf{1 + 3 = 4}

\mathsf{8 + 4 = 12}

\boxed{\boxed{\large\text{Answer: \bf \huge12 }}}\huge\checkmark

\text{Good luck on your assignment and enjoy your day! }

~\frak{Amphitrite1040:)}

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