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Zepler [3.9K]
4 years ago
5

There is 100 people in each train car so how many people are they in each train there is ten cars in each train and thirty train

s help me please
Mathematics
1 answer:
snow_tiger [21]4 years ago
6 0

Information given:

- <u>100 passengers in each train car</u>

- <u>10 train cars total</u>

- <u>30 trains total</u>

<em>So we are trying to find the total number of passengers on all 30 trains?</em>

<u />

Step-by-step:

100 x 10 = <u>1,000 passengers per train</u>

1,000 x 30 = <u>30,000 passengers total within the 30 trains</u>

Answer: <u>30,000 passengers</u>

<em>Hopefully this answered your question :)</em>

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The price of a car is usally £12,500 it is reduce to £11,625 what is the percentage of the reduction
irina [24]

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What is .25 written in its simplest fraction form? A. 2/5 B. 4/1 C. 1/4 D. 100/25
Furkat [3]
0.25 = 25/100

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3 years ago
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Can anyone help me with this please.<br> I’ll mark you as a brainliest.
seropon [69]

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x = 21

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A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

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-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

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t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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