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Nimfa-mama [501]
2 years ago
11

students were grouped into 4 classrooms with an average of 33.5 students per classroom. if the students were regrouped into 5 cl

assrooms, what would be the average number of students in each room?
Mathematics
1 answer:
erma4kov [3.2K]2 years ago
4 0

Answer:

26.8

Step-by-step explanation:

Because 33.5 was the average of how many students were in a classroom (if there were 4 classrooms), then 33.5 is can somewhat be the average of how many students were in one classroom; 1/4 of the classrooms can hold an average of 33.5.

Total students that were divided into 4 classrooms :

33.5 × 4 = 134

So if there is 134 students that will be grouped into 5 different classrooms, divide 134 by 5 to find the average amount of students that will be grouped.

134 ÷ 5 = 26.8

So, the average amount of students that will be grouped if they were divided into 5 classrooms would be 26.8 .

Hope this helped !

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Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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