<p>Does anyone know how to do number 2 on this pdf.</p>
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<p>I don’t know if this is right but
a) take the integral of the rate function L(t) from 0 to 18
b) since the graph is over the y value of 150 from 13 to 16 i guess, i think you’d have to be able to draw the lines up to know for sure, you find the average value from 13 to 16, so 1/(16-13) times the integral from 13 to 16 of the rate function
c) since there are 500 cars going straight through the max number of cars going left for there to be no signal would be 400. but you can see that at hours 14 and 15, there are more than 200 cars turning left, so that’s a total of over 400. so the intersection does need a signal</p>
<p>Thats how I’ll do too…</p>
<p>or well if it’s a calculator problem, for part b you could graph it to see exactly where it hits 150</p>
<p>is this what you mean</p>
<p>a) 18 to 0 L(t) dt</p>
<p>b) f(16)-f(13)/16-13</p>
<p>well the first one is 0 to 18, not 18 to 0
the second one, i thought it would be using the formula for the avg value of a function… 1/(b-a) times the integral of L(t) dt from a to b, where a is 13 and b is 16. i’m not sure though, maybe someone else knows for sure?</p>