please answer this fast. why is nobody answering?

Question

t (minutes)--> 0 1 2 4 6 7

W(t) (ounces)--> 0 8 13 22 25 33

water is dripping into a cup. The amount of water in the cup at time t, 0 less than or equal to t which is less than or equal to 6, is given by the differentiable function W, where t is measured in minutes. Selected values of W(t), measured in ounces are given in the table above.

a) use the data in the table to approximate W'(5). Show the computation that led you to your answer and interpret your answer in the context of the problem using proper units. show all calculus work.

b) for 0 less than or equal to t less than or equal to 7, must there be a time when W9t) =18? Explain

c) for t>7, water leaks out of the cup at a constant rate of 0.2 ounces per minute. At what time will there be no water left in the cup?

Answer 1

t (minutes)--> 0 1 2 4 6 7

W(t) (ounces)--> 0 8 13 22 25 33

33/7 = avergage rate in, W(t)=(33/7)t ,,,,,,,,0≤t≤7

the pattern is a bit random so best to use linear model

W(5) approximately 5*33/7

W(t) is a always increasing differentialable (implying continuity and intermediate value theorem),

therefore at some point between of the cup having zero and having 33 there must be 18.

.2 oz/min *x=33

x=5*33