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Answer:
(a) As time increases, the amount of water in the pool increases.
11 gallons per minute
(b) 65 gallons
Step-by-step explanation:
From inspection of the table, we can see that <u>as time increases, the amount of water in the pool increases</u>.
We are told that Ann adds water at a constant rate. Therefore, this can be modeled as a linear function.
The rate at which the water is increasing is the <em>rate of change</em> (which is also the <em>slope </em>of a linear function).
Choose 2 ordered pairs from the table:
Input these into the slope formula:
Therefore, the rate at which the water in the pool is increasing is:
<u>11 gallons per minute</u>
To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:
When Ann had added no water, x = 0. Therefore,
So there was <u>65 gallons</u> of water in the pool before Ann starting adding water.
Answer:
Step-by-step explanation:
Answer: $ 55
Step-by-step explanation:
When interest is compounded continuously, the final amount will be
When interest is compounded daily, the final amount will be
, where P= Principal , r = rate of interest , t = time
For Hunter , P= $750, r =
t = 18 years
For London , P= $750, r =
t = 18 years
Difference = $ 2471.48 - $ 2416.24 =$ 55.24≈$ 55
Hence, Hunter would have $ 55 more than London in his account .