Question

The price of Stock A at 9 A.M. was ​$15.03. Since​ then, the price has been increasing at the rate of ​$0.05 each hour. At noon the price of Stock B was ​$15.53. It begins to decrease at the rate of ​$
0.14 each hour. If the two rates​ continue, in how many hours will the prices of the two stocks be the​ same?

Answer 1

After 1.842 hours the price of Stock A and Stock B will be the same using a set of linear equations.

A grouping of one or more linear equations containing the same variables is known as a system of linear equations.

For Stock A :

The initial price = $15.03

The rate of increase = $0.05

For Stock B :

The rate of decrease = $15.53

The rate of decrease per hour = $0.13

Then the price of stock A at noon will be:

$ 15.03 + $ (0.05 × 3) = $ 15.03 + $ 0.15 = $15.18

The period of time during which the stock price will remain constant:

15.18 + 0.05t = 15.53 - 0.14t

Simplifying the terms we get,

0.05t + 0.14t = 15.53 - 15.18

0.19t = 0.35

t = (0.35 ÷ 0.19) hours

t = 1.842 hours

Therefore, the stocks will be of equal price after 1.842 hours.

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