1) Since this is a Continuously Compounded operation in a 10 yrs period, then we can write out the following equation:
2) Plugging into the equation the given data and since Otto is 20 yrs old and he plans to get $4,000 in ten years, we can write out:
3) Thus the rate Otto needs is
Note that since the 0.1386 the six here is greater than 5 then we can round up to the next thousandth, in this case: 0.139. For the 0.1386 is closer to 0.139 (0.004) than to 0.138 (0.006).
Or 13.9%
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.