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3 years ago

48% of the students in the class play sports. 12 students in the class play sports , how many students are in the class?

Mathematics

2 answers:

Setler [38]3 years ago

3 0

The answer is 25 students.

Novay_Z [31]3 years ago

3 0

This problem is quite difficult if you don't remember a technique to solving it.
I will help teach you this tactic.
First, set the percentage over 100, which is 48/100 as a fraction (which is what "over 100" means).
Now, put 12 over x, which is the total amount of students in the class. It should look like this: 12/x.
Now that we've done this, set them up as a proportion.
Your proportion should be:
12/x = 48/100
This is where it becomes a little confusing if you do not remember this correctly.
Cross multiply the denominator (the bottom number of a fraction) of 12/x by the numerator (the top number of a fraction) of 48/100.
This number should be:
48x.
Now, cross multiply the denominator of 48/100 by the numerator of 12/x.
This number should be:
1200.
Our equation now is:
48x = 1200.
To solve for this, divide both sides by 48 to solve for x.
48x / 48 = x
1200 / 48 = 25.
Your equation and solved problem is:
There are 25 total students.
I hope this taught you something and I hope it's very helpful!

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1. If the length of a rectangle is 6 more than its width, and its perimeter is 48

Salsk061 [2.6K]

For this case we have that by definition, the perimeter of the rectangle is given by:

$P = 2w + 2L$

Where:

W: Is the width of the rectangle

L: is the length of the rectangle

According to the data we have:

$L = 6 + w\\P = 48$

Substituting:

$48 = 2w + 2 (6 + w)\\48 = 2w + 12 + 2w\\48 = 4w + 12\\48-12 = 4w\\36 = 4w\\w = \frac {36} {4}\\w = 9$

So, the width of the rectangle is 9 inches

$L = 6 + 9 = 15$

So, the length of the rectangle is 15 inches

Answer:

the width of the rectangle is 9 inches

the length of the rectangle is 15 inches

5 0

2 years ago

Which of the following represents the difference between the central angles of the above regular polygons? Round to the nearest

solong [7]

Answer:

its 15.43

Step-by-step explanation:

you didnt put a picture for the polygons but im pretty sure this is from math nation.

6 0

2 years ago

Tuto

alexgriva [62]

answer is 40%

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8 0

1 year ago

Prove that if a and b are positive integers whose sum is a prime p, their greatest common divisor is 1.

Tasya [4]

Proof by contadiction.

Let assume there exist such positive integers $a$ and $b$ whose sum is a prime number $p$ , that their greatest common divisor is greater than 1.

$a,b\in\mathbb{Z^+}\\d=\text{gcd}(a,b)>1\\\\a=de\\b=df\\e,f\in\mathbb{Z^+}\\\\a+b=p\\de+df=p\\d(e+f)=p\\$

Since $p$ is a prime number and $d>1$ , then $d=p \wedge e+f=1$ , but we assumed earlier that $e,f\in\mathbb{Z^+}$ , and there are no two positive integers that sum up to 1.