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Goryan [66]
2 years ago
9

Mrs. Bendit’s 1st Block class has 28 students. There are 7 girls in the class. What percent of boys are in the class?

Mathematics
2 answers:
Gemiola [76]2 years ago
7 0

Answer:

Boys: 75%

Girls: 25%

Step-by-step explanation:

Total: 28 Students: 100%

7 Girls in clas

28(total)-7 (girls)= 21 Boys in class

__________________

If we divide the amount  boys OR girls ( depending on which gender) after the toal and move the decimal to the right twice after dividing) we will get our answer.

Formula:

(Amount of Boy or girls) = x

X/28

So since we are trying to get the percentages of boys, we will replace X with the amount of boys

21/28 = .75 ( then move the decimal to the right TWICE) =  75  

Then you should get your percentages: 75%  Also if your lookings for girls percentages, you basically subtract 75 ( boys percentages) with the total percentages ( 100) then you should get your answer for the girls: 25%

Masteriza [31]2 years ago
7 0

Answer: 75%

Step-by-step explanation:

First, the boys in the class are21 since 28-7 is 21,

So, since our denominator in 21/28 is 28, we could adjust the fraction to make the denominator 100. To do that, we divide 100 by the denominator:

100 ÷ 28 = 3.5714285714286

Once we have that, we can multiply both the numerator and denominator by this multiple:

\frac{21 x 3.5714285714286}{28 x 3.5714285714286} =\frac{75}{100}

Now we can see that our fraction is 75/100, which means that 21/28 as a percentage is 75%.

We can also work this out in a simpler way by first converting the fraction 21/28 to a decimal. To do that, we simply divide the numerator by the denominator:

21/28 = 0.75

Once we have the answer to that division, we can multiply the answer by 100 to make it a percentage:

0.75 x 100 = 75%

If you like mark as brain list, Please.

Hope this help :)

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What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

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