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Juliette [100K]
3 years ago
10

A train traveled 720km in 9h. How far would it travel in 11 hr? How long would it take to go 1120km?

Mathematics
2 answers:
kompoz [17]3 years ago
6 0
It would go 880 km in 11 hours, and it would take 14 hours to go 1120 km
Ahat [919]3 years ago
3 0
How far would it travel in 11 hours?
First, we have to find how much the train travels in 1 hour. We are finding the unit rate. So, divide 720 by 9 and we get 80. The train travels 80 km in one hour. We have to find how much it travels in 11 hours. We multiply 80 by 11 and we get 880.
<u>The train would travel 880 km in 11 hours.</u>

How long would it take the train to travel 1,120 km?
Since we know that the train can travel 80 km per hour , we can divide 1,120 by 80. When we divide we get 14.
<u>The train would travel 1,120 km in 14 hours.</u>

-Hope this helped! :)
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The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
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<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

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y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

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