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Ber [7]
3 years ago
9

Oliver interviewed 30% of the 9th grade class and 70% of the 10th grade class at his school. Jenny interviewed 75% of the 9th gr

ade class and 25% of the 10th grade class at the same school. Oliver interviewed a total of 176 students and Jenny interviewed 140 students.
How many more 10th graders than 9th graders were interviewed?
A
36

B
80

C
120

D
200
Mathematics
1 answer:
mixer [17]3 years ago
5 0

Answer:

A . 36

Step-by-step explanation:

We are given a total of 176 interviewed by Oliver and a total of 140 interviewed by Jenny. To find how many more 10th graders than 9th graders were interviewed, subtract the totals given

176 - 140 = 36

This is how we came to the answer:

We are given 70% of the 10th-grade and 30% of the 9th-grade with a total of 176 for Oliver.

While we're given 75% of the 9th-grade class and 25% of the 10th-grade with a total of 140 interviewed by Jenny

<u>Oliver's Interviewees</u>

  • 10-graders

Firstly, let's find what the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

70% of 176 =

\frac{70}{100} * \frac{176}{1}

Cross multiply

123.2 were 10-graders interviewed by Oliver

  • 9th-graders

Now, to find the number of 9th-graders was interviewed by Oliver; find the percentage of the 9th-graders by the total;

30% of 176 =

\frac{30}{100} * \frac{176}{1}

Cross multiply

52.8 were 9th-graders interviewed by Oliver

<u>Jenny's Interviewees</u>

  • 9th-graders

Firstly, let's find what the number of 9th-graders was interviewed by Jenney; find the percentage of the 9th-graders by the total;

75% of 140 =

\frac{75}{100} * \frac{140}{1}

Cross multiply

105 students were 9th-graders interviewed by Jenney.

  • 10th-graders

Now, to find the number of 10th-graders was interviewed by Jenney; find the percentage of the 10th-graders by the total;

25% of 140 =

\frac{25}{100} * \frac{140}{1}

Cross multiply

35 students were 10th-graders interviewed by Jenney.

<u />

<u>Total calculation</u>

Use the results and sum them up by 9th-grade plus 9th-grade and 10th-grade plus 10-grade. Then subtract the amount gotten from 9th-grade away from the amount gotten from 10th-grade;

Oliver's 9th-grade = 52.8

Jenny's 9th-grade = 105

105 + 52.8 = 157.8

Oliver's 10th-grade = 123.2

Jenny's 10th-grade = 35

123.2 + 35 = 158.2

Total calculation: 158. 2 - 157.8 = 0.4

<h2>Therefore, there are 36 more 10th than 9th.</h2>

<u />

<h3><u>Extra Info:</u></h3>

<u>Oliver's Interviewees Percentage</u>

Since we are given 30% of the 9th-grade class and 70% of the 10th-grade class, first, let's add the percentages. To do so, set it up as a fraction;

30% = \frac{30}{100} while, 70% = \frac{70}{100}

Now solve it;

\frac{30}{100} + \frac{70}{100}

Simplify; cancel bottom zero's;

\frac{30}{1} + \frac{70}{1}

Add the remaining numerators;

30 + 70 = 100

Which is 100%

<u>Jenny's Interviewees Percentage</u>

Since we're given 75% of the 9th-grade class and 25% of the 10th-grade, it will end up the same answer. I'll show you how; first, let's add the percentages. To do so, set it up as a fraction;

25% = \frac{25}{100} and, 75% = \frac{75}{100}

Now solve it;

\frac{25}{100} + \frac{75}{100}

Simplify; cancel bottom zero's

\frac{25}{1} + \frac{75}{1}

Add the remaining numerators;

25 + 75 = 100

Meaning 100%

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Ms. Lady works-out with Ms. Beetle; her average crawling time has increased to 4 feet per minute, while Ms. Beetle, who is used
ira [324]

Answer:

Senn needs to give Ms. Lady 4.6 minutes  head start

Senn needs to give Ms. Beetle 18.4 minutes  head start

The equation for each bug are;

For Senn, the equation for the experiment is 5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is 4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is 4 × (t₁ + 18.4) = 92 feet

Please find attached the required graph and table of values

Step-by-step explanation:

The given parameters are;

The average crawling time of Ms. Lady = 4 feet per minute

The average crawling time of Ms. Beetle = 2.5 feet per minute

The average crawling time of Senn = 5 feet per minute

The distance to the rose bush = 92 feet

Therefore, we have;

The time, t, duration for Senn to arrive at the Rose bush is given by the following relation,

Time, t = Distance, d/(Speed, s)

Given that the speed of the bugs is equal to their average crawling time, we have

For Senn

Time, t = 92/(5 ft/Min) = 18.4 minutes

For, Ms. Lady

Time, t = 92/(4 ft/Min) = 23 minutes

For, Ms. Beetle

t = 92/(2.5 ft/Min) = 36.8 minutes

Therefore;

Senn needs to give Ms. Lady 23 - 18.4 = 4.6 minutes head start

Similarly, Senn needs to give Ms. Beetle 36.8 - 18.4 = 18.4 minutes  head start

The equation for each bug are therefore;

For Senn, the equation for the experiment is given as follows ;

5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is given as follows;

4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is given as follows;

4 × (t₁ + 18.4) = 92 feet.

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