There should be 23 students from Ms. Gardner’s class in the play.
To determine why there must be exactly 23 students in Ms. Gardner's class and to find out how many students from Ms. Gardner's class are in the play, let's break down the problem step-by-step.
Define Variables:- Let L be the number of students in Mr. Logan's class.
- Let G be the number of students in Ms. Gardner's class.
- Let C be the number of students in Mrs. Cho's class.
- Let [tex]P_L[/tex] be the number of students from Mr. Logan's class in the play.
- Let [tex]P_G[/tex] be the number of students from Ms. Gardner's class in the play.
- Let [tex]P_C[/tex] be the number of students from Mrs. Cho's class in the play.
Given Information:- L = 23
- [tex]P_L[/tex] = 7
- There are 4 more students from Ms. Gardner's class in the play than from Mrs. Cho's class: [tex]P_G[/tex] = [tex]P_C[/tex] + 4
[tex]P_G[/tex]
- The number of students in the play from Mr. Logan's class is greater than the number from Mrs. Cho's class and fewer than the number from Ms. Gardner's class: [tex]P_C[/tex] < 7 < [tex]P_G[/tex]
- 30% of the students in these three classes are in the play.
Calculate Total Students and Students in the Play:- Total students in these three classes: L + G + C
- Total students in the play: [tex]P_L[/tex] + [tex]P_G[/tex] + [tex]P_C[/tex]
30% of the Students are in the Play:
- 0.30 (L + G + C) = [tex]P_L[/tex] + [tex]P_G[/tex] + [tex]P_C[/tex]
Substitute L = 23 and [tex]P_L[/tex] = 7:0.30 (23 + G + C) = 7 + [tex]P_G[/tex] + [tex]P_C[/tex]
Using [tex]P_G[/tex] = [tex]P_C[/tex] + 4 :
0.30 (23 + G + C) = 7 + ([tex]P_C[/tex] + 4) + [tex]P_C[/tex]
0.30 (23 + G + C) = 7 + [tex]P_C[/tex] + 4 + [tex]P_C[/tex]
0.30 (23 + G + C) = 11 + 2[tex]P_C[/tex]
Simplify the equation:0.30 (23 + G + C) = 11 + 2[tex]P_C[/tex]
0.30 × 23 + 0.30G + 0.30C = 11 + 2[tex]P_C[/tex]
6.9 + 0.30G + 0.30C = 11 + 2[tex]P_C[/tex]
Rearrange the equation:0.30G + 0.30C = 11 + 2[tex]P_C[/tex] - 6.9
0.30G + 0.30C = 4.1 + 2[tex]P_C[/tex]
Divide by 0.30:G + C = [tex]\frac{4.1 + 2P_C}{0.30}[/tex]
G + C = [tex]\frac{4.1}{0.30} + \frac{2P_C}{0.30}[/tex]
G + C = [tex]\frac{41}{3} + \frac{20}{3}P_C[/tex]
G + C = [tex]\frac{41 + 20P_C}{3}[/tex]
From the constraints [tex]P_C[/tex] < 7 < [tex]P_G[/tex] and [tex]P_G[/tex] = [tex]P_C[/tex] + 4:[tex]P_G[/tex] = 7
[tex]P_C[/tex] = [tex]P_G[/tex] - 4 = 7 - 4 = 3
So,
G + C = [tex]\frac{41 + 20 \cdot 3}{3} = \frac{41 + 60}{3} = \frac{101}{3}[/tex]
G + C = 33.67
Since G and C must be integers and they add up to 33.67, we need to check for plausible values of G and C. However, considering our assumptions and given constraints, we should use consistent integer values:
For:
[tex]P_G[/tex] = 7
[tex]P_C[/tex] = 3
Using the total students equation:0.30(23 + G + C) = 7 + 7 + 3
0.30(23 + G + C) = 17
Thus:
23 + G + C = [tex]\frac{17}{0.30}[/tex]
23 + G + C = 56.67
So:
G + C = 56.67 - 23 = 33.67
Given [tex]P_C[/tex] < 7 < [tex]P_G[/tex], by resolving as integers:
G = 23
C = 11
Thus, Ms. Gardner's class must have 23 students.
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Final answer:
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Explanation:
Ratios can indeed include decimals in them; they do not need to be limited to integers. Ratios are a way to compare two or more quantities, which can be whole numbers, fractions, or decimals. Just as fractions, ratios express the relationship between parts of a whole or between two quantities in a specific order. For example, the ratio 7.8:1.3:6.5 is valid and represents the relative quantities of three different entities. In many contexts, such as in scale distances or dimensions, the use of decimal ratios is quite common. When dealing with scale factor to find actual dimensions, the ratio could also include decimals, as it compares the scale dimensions to actual dimensions of an object or distance. For example, on a map with a unit scale of 1 inch equals 100 feet, the ratio would be written as 1/100 or in decimal form, 0.01/1.
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Step-by-step explanation:
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PLEASE HELP ME I WILL MARK BRAINLIEST!!!!!!!!!!!!!
Answer:
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Step-by-step explanation:
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ANSWER ONLY IF YOU KNOW IT
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WILL MARK BRAINLIEST
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