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svetlana [45]
3 years ago
15

The four Fifth Grade classes at Smith Elementary School are having a competition to collect Box Tops. Below is the information a

bout the number of Box Tops that each class has collected.
• Mrs. Jacoby’s class: 240 box tops
• Mrs. Albert’s class: 2/3 as much as Mrs. Jacoby’s class
• Mrs. Thomas’ class: 1 and 1/3 as much as Mrs. Jacoby’s class
• Mr. Williams’ class: 1 and 2/3 as much as Mrs. Jacoby’s class
• Mrs. Yount’s class: 5/6 as much as Mrs. Jacoby’s class.


Part 1:
Which class collected more Box Tops than Mrs. Jacoby’s class? How do you know?



Without calculating, list the classes in descending order in terms of the number of box tops.



Write an explanation about how you found the order of the classes.




Part 2:
Find the actual number of box tops for each class.







Write an explanation about how you found your answers.
Mathematics
1 answer:
Rzqust [24]3 years ago
7 0

Answer:

ysgwhdbjsbamssnkfnenxndnkx4iihug58hs

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,
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<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

<h2>Why?</h2>

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

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