Mean= 4.75
Median= 4
Mode= 3,7, and 2
Log(4k - 5) = log(2k - 1)
log(4k) - log(5) = log(2k) - log(1)
0.6020599913k - 0.6989700043 = 0.3010299957k - 0
0.6020599913k - 0.6989700043 = 0.3010299957k
<u>-0.6020599913k -0.6020599913k</u>
<u>-0.6989700043</u> = <u>-0.3010299957k</u>
-0.3010299957 -0.3010299957
2.321928094 = k
The number of gallons of water in the tank at t=10 is
... W(10) = 160,000 -10(8000 -10) = 80100
The number of gallons of water in the tank at t=10.5 is
... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25
The rate of change over the interval is
... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5
The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.
The sign is negative, so the amount of water is decreasing.
Answer:
a = 70 km/h - 20 km/h / 0.008 h = 50 km/h / 0.008 h = 6250 km/h²