Maths please someone help me out!!!

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3. The efficiency of an industrial pump decreases by a fixed percentage each year. The percentage decrease each year is with respect to the efficiency of the pump at the start of each year. Over each four year period, the efficiency decreases by 80%. Let us write e = f(t) so that the efficiency e of the pump, as a percentage of it’s initial efficiency, is a function of the number of years t since the pump has been in operation.

(a) Make a table of values of the function for changes in t of 4, starting from t = 0.

(b) Explain why the formula is given by e = f(t) = 100(0.2)t/4.

(c) By a process of guessing, checking and fixing (no logarithms yet!) estimate the time taken for the efficiency to reach 50%. Choose a reasonable level of accuracy.

(d) Show that the halving time is in fact constant by showing that no matter what time we consider, say t = a, if you then add the time found in the previous part, the efficiency is halved.

(e) Use the halving time to make a sketch of the function over an 8 year period.

  • Maths please someone help me out!!! -

    .2^(t/4) multiplies by 0.2 every time t increases by 4. That's how the efficiency is described in words.

    so, get out your calculator (online or physical) and evaluate some values for t. As a sanity check, 50% is a little more than twice as much as 20%, so expect t to be somewhere around half of the 4-year cycle.

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