Mathematics: The strength of students in a school when increased by 8% becomes 2160. find the original strength of students in the school.

The strength of students in a school when increased

Ответ:

layla07

10.03.2022

1987.2

Step-by-step explanation:

We can write a proportion to find the total amount in the choir using the information given. A proportion is two equivalent ratios set equal to each other. Since we know the strength of students increased by 8% to 2160 then the original was 92%.

$\frac{92}{100}=\frac{y}{2160}$

We will cross multiply the numerator of one ratio with denominator of the other. And then solve for y.

92(2160)=100(y)

198720=100y

y=1987.2

There are 1987.2 as the original strength of students.

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Ответ:

jinn53

22.11.2022

Mathematics

4 is the GCF of 16s^3t , 40s^5 and 68t^2

Step-by-step explanation:

Find the GCF of 16s^3t , 40s^5 and 68t^2 .

GCF is the largest number that divide these numbers.

List the prime factor of 16s^3t , 40s^5 and 68t^2 as:

$16s^3t = 2 \cdot 2 \cdot 2 \cdot 2 \cdot s \cdot s \cdot s \cdot t$

$40s^5 = 2 \cdot 2 \cdot 2 \cdot 5 \cdot s \cdot s \cdot s \cdot s \cdot s$

$68t^2 = 2 \cdot 2 \cdot 17 \cdot t \cdot t$

Common factor of 16s^3t , 40s^5 and 68t^2 is, $2 \cdot 2= 4$

Greatest common factor of 16s^3t , 40s^5 and 68t^2 is, 4

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The strength of students in a school when increased by 8% becomes 2160. find the original strength of students in the school.

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