Answer:
T = 40 (minutes?)
Step-by-step explanation:
Given the equation G = 50 + 20T, where G = the total amount of gas and T = the amount of Time, we know that there is already 50 gallons of gas in the tank and that the rate at which it is pumped is 20 gallons per minute (I am assuming, though it is not stated in the problem). Given that the tank holds 850 gallons, we can plug in the values into the equation and solve for the missing variable, T:
850 = 50 + 2T
Subtract 50 from both sides: 850 - 50 = 50 + 2T - 50 or 800 = 2T
Divide 2 from both sides: 800/2 = 2T/2
Solve for T: 40 = T
4,000,000,39
there you go!
Answer:
.8
Step-by-step explanation:
S= studies for
B= score of b or higher
we want P(S|B)
Using Bayes' theorem we can solve for the probability

9514 1404 393
Answer:
- maximum: 15∛5 ≈ 25.6496392002
- minimum: 0
Step-by-step explanation:
The minimum will be found at the ends of the interval, where f(t) = 0.
The maximum is found in the middle of the interval, where f'(t) = 0.
![f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)](https://tex.z-dn.net/?f=f%28t%29%3D%5Csqrt%5B3%5D%7Bt%7D%2820-t%29%5C%5C%5C%5Cf%27%28t%29%3D%5Cdfrac%7B20-t%7D%7B3%5Csqrt%5B3%5D%7Bt%5E2%7D%7D-%5Csqrt%5B3%5D%7Bt%7D%3D%5Csqrt%5B3%5D%7Bt%7D%5Cleft%28%5Cdfrac%7B4%285-t%29%7D%7B3t%7D%5Cright%29)
This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...
f(5) = (∛5)(20-5) = 15∛5
The absolute maximum on the interval is 15∛5 at t=5.