Amount of water in the pool at the end of the day is 5457798.7 gallons
<u>Explanation:</u>
Given:
Initial amount of water in the pool = 45,000 gallons
Increase in amount = 0.75 in per minute
Time, t = 1 day
t = 24 X 60 min
t = 1440 min
So,
Increase in amount of water in 1 day = 0.75 in X 1440
= 1080 in
Volume of 1080 in of water = (1080 in)³
Volume from cubic inch to gallon = 5453298.7 gallon
Amount of water in the pool at the end of the day = 45000 + 5453298.7 gallon
= 5457798.7 gallon
Answer:
f(x) is going to be on the y axis or intercept and what number it is on the line is in quotes f(x)="6"+2(1.05)^x it would be at 6 on the y axis.
0.41 > 3/7 > 43% this is the answer
You can see how this works by thinking through what's going on.
In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97
What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.
So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.
So the population after x years is 1200 * 0.97x
