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denpristay [2]
3 years ago
9

 The price of Stock A at 9 A.M. was ​$13.56. Since​ then, the price has been increasing at the rate of ​$0.12 each hour. At noon

the price of Stock B was ​$14.31. It begins to decrease at the rate of ​$0.14 each hour. If the two rates​ continue, in how many hours will the prices of the two stocks be the​ same?
Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
6 0

Answer:

The prices of the two stocks be the​ same after 2.88 hours after 12 pm

Step-by-step explanation:

Price of stock A at 9 a.m.= $13.56

The price has been increasing at the rate of ​$0.12 each hour.

Increase in price after 3 hours =0.12 \times 3 =0.36

Price of Stock A at noon= 13.56+0.36 = $13.92

Let n be the no. of hours after which the prices of both the stocks will be same .

So, Increase in price after n hours = 0.12 n

Price of Stock A after n hours = 13.92+0.12 n

Price of stock B = $14.31

The price has been decreasing at the rate of ​$0.14 each hour.

Let n be the no. of hours

So, Increase in price after n hours = 0.14 n

Price of Stock B after n hours =14.31-0.14 n

ATQ

13.56+0.12 n= 14.31-0.14 n

-13.56+14.31=0.14 n + 0.12 n

0.75=0.26 n

\frac{0.75}{0.26}=n

2.88=n

Hence the prices of the two stocks be the​ same after 2.88 hours after 12 pm .

Radda [10]3 years ago
4 0
$6.36 would be your answer!
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