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Anestetic [448]
3 years ago
7

Two trains leave stations 252 miles apart at the same time and travel toward each other. One train travels at 80 miles per hour

while the other travels at 100 miles per hour. How long will it take for the two trains to meet?
Mathematics
1 answer:
Svetlanka [38]3 years ago
4 0
First things first, distance = speed x time
So let's have "t" represent the time by both trains.
The first train shall be Train A and the other becoming Train B, so now we have Train A going 80t miles and Train B going 100t miles.


If two trains head toward each other, they will have traveled a total of 252 miles. The faster train will have traveled further (Train B), so the meeting point will be closer to the slower train's starting point (Train A).

So now we do 80t + 100t = 252
Which becomes 180t = 252
And by diving 252 by 180, you'll get that t is 1.4
This would mean they would meet at one hour and 24 minutes.
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4 years ago
The following are the number of birth per year per 1,000 population for 20 countries: 34, 24, 10, 15, 22, 15, 17, 22, 10, 17, 25
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Answer:

A. D = None of the above

B. B = 18

C. A = 15

D. D = None of the above

Step-by-step explanation:

Rearranging the population from the highest to the lowest.

10, 10, 12, 15, 15, 15, 15, 17, 17, 18, 18, 20, 22, 22, 24, 25, 31, 32, 34, 37

Mean = (summation of all the 20 samples)/ no of samples

Mean per year per 1000 population = 409/20

= 20.45

Median = the middle value (for odd numbered samples)

= the sum of the middle value ÷ 2 ( for even numbered samples)

Median birth per year per 1000 population = (18 + 18)/2

= 18

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6 0
3 years ago
Answer this question
Murljashka [212]

Answer:

a = p * q

b = p * s + q * r

c = r * s

Step-by-step explanation:

In the trinomial ax² + bx + c

a is the coefficient of x²

b is the coefficient of x

c is the numerical term

∵ The trinomial is ax² + bx + c

∵ Its factors are (px + r) and (qx + s)

∴ ax² + bx + c = (px + r)(qx + s)

∵ (px + r)(qx + s) = (px)(qx) + (px)(s) + r(qx) + (r)(s)

∴ (px + r)(qx + s) = pqx² + (psx + qrx) + rs

∴ ax² + bx + c = pqx² + (ps + qr)x + rs

→ By comparing the two sides

∵ ax² = pqx² ⇒ divide both sides by x²

∴ a = pq

∵ bx = (ps + qr)x ⇒ Divide both sides by x

∴ b = ps + qr

∴ c = rs

∴ a = p * q

∴ b = p * s + q * r

∴ c = r * s

5 0
3 years ago
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