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2 years ago

9

In a classroom,there are 12 girls and 8 boys.What fraction of the students in class are girls?What percent of the class are girl

s?

1 answer:

tekilochka [14]2 years ago

4 0

Answer:

Step-by-step explanation:

Add the total number of students:

Boys: 8, + Girls: 12 = 20 students

$\frac{12}{20} = \frac{6}{10}$ of the students are girls

6/10 also equates to 60%, which is the percent of girls in the class.

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shepuryov [24]

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$21.44+$5.36=$26.80

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2 years ago

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A certain genetic condition affects 8% of the population in a city of 10,000. Suppose there is a test for the condition that has

enot [183]

Solution:

Population in the city= 10,000

As genetic condition affects 8% of the population.

8 % of 10,000

$=\frac{8}{100}\times 10,000=800$

As, it is also given that, there is an error rate of 1% for condition (i.e., 1% false negatives and 1% false positives).

So, 1% false negatives means out of 800 tested who are found affected , means there are chances that 1% who was found affected are not affected at all.

So, 1% of 800 $=\frac{1}{100}\times 800=8$

Also, 1% false positives means out of 10,000 tested,[10,000-800= 9200] who are found not affected , means there are chances that 1% who was found not affected can be affected also.

So, 1% of 9200 $=\frac{1}{100}\times 9200=92$

1. Has condition Does not have condition totals = 800

2. Test positive =92

3. Test negative =8

4. Total =800 +92 +8=900

5. Probability (as a percentage) that a person has the condition if he or she tests positive= As 8% are found positive among 10,000 means 9200 are not found affected.But there are chances that out of 9200 , 1% may be affected

$=\frac{\text{1 percent of 9200}}{9200}\\\\ \frac{\frac{1}{100}\times 9200}{9200}=\frac{92}{9200}\\\\ =0.01$

that is Probability equal to 0.01 or 1%.

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2 years ago

Anne and Nancy use a metal alloy that is 20% copper to make jewelry. How many ounces of a 14% and 24% alloy must be mixed to for

Gala2k [10]

<span>copper +copper = copper
0.14x+0.24(100-x) = 0.20*100 = 20</span>

7 0

2 years ago

Will give brainliest if answer correct// if p(x)= 3x^6 - 5x^5 - x + 7 , then p(2) equals: ??

djverab [1.8K]

$\\ \tt\hookrightarrow p(2)=3(2)^6-5(2)^5-2+7$

$\\ \tt\hookrightarrow p(2)=3(64)-5(32)+5$

$\\ \tt\hookrightarrow p(2)=192-160+5$

$\\ \tt\hookrightarrow p(2)=32+5$

$\\ \tt\hookrightarrow p(2)=37$

Option B is correct

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2 years ago

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How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

___

<h3>9b.</h3>

$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

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<h3>10b.</h3>

$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$

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2 years ago