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AleksandrR [38]
3 years ago
14

The ratio of men to women on the First-Year

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

9

Step-by-step explanation:

the total ration is 8

the number of women = (3/8) x 24

                                     = 9

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What is the coefficient of the expression:<br> 3n+4
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Answer:

The coefficient is 3, the number right before the variable.

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Help me please solve the circled question show your work
Rasek [7]

<u>Part 1)</u> Use the fact that there are 12 inches in a foot

a) How many inches tall is a 7 foot basketball player?

we know that

1 foot=12 inches

so by proportion

\frac{12}{1} \frac{inches}{foot} =\frac{x}{7} \frac{inches}{feet} \\ \\x=12*7\\ \\x=84\ inches

therefore

<u>the answer Part 1a) is</u>

84\ inches

Part b) If a yard is a 3 feet long, how many inches are in a yard?

we know that

1 yard=3 feet

1 foot=12 inches

so

Convert 3 feet to inches

by proportion

\frac{12}{1} \frac{inches}{foot} =\frac{x}{3} \frac{inches}{feet} \\ \\x=12*3\\ \\x=36\ inches    

therefore

<u>the answer Part 1b) is</u>

36\ inches

<u>Part 2) </u>At the farmer's market two pounds of peaches cost $4.20 How much will five pounds cost?

by proportion

\frac{4.20}{2} \frac{\$}{pounds} =\frac{x}{5} \frac{\$}{pounds} \\ \\2*x=4.2*5\\ \\x=4.2*5/2=\$10.5    

therefore

<u>the answer Part 2) is</u>

\$10.5

<u>Part 3) </u>Janice mother gave her a ten dollars bill to buy five pounds each of bananas and apples at the grocery store. When she got there she found that bananas were 80 c per pound and apples were $1.40 per pound.

Did Janice's mother give her enough money? If so, should she receive any change? If not, how much more money does she need?

 we know that

<u>Find the cost of the bananas</u>

5\ pounds*0.80\frac{\$}{pounds}=\$4

<u>Find the cost of the apples</u>

5\ pounds*1.40\frac{\$}{pounds}=\$7

Sum the cost of the bananas and the cost of the apples

\$4+\$7=\$11

\$11 > \$10------> Janice's mother didn't give her enough money

She needs more money

She needs-------> \$11-\$10=\$1

therefore

<u>the answer part 3) is</u>

She needs  \$1 more


3 0
3 years ago
What numbers are greater than 20 that has 9 as a factor
OLga [1]
Any of 9's multiples besides 9 and 18 will work.

So: 27, 35, 45, 54 ,63, 72, 81, 90, 99, 108, 117, 126, 135, and so on
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Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum
lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\

\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\  (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\

=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3}  \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\

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2 years ago
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