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December 20, 2014

December 20, 2014

Posted by **jenny** on Tuesday, October 14, 2008 at 9:59pm.

- college algebra -
**Reiny**, Wednesday, October 15, 2008 at 9:09amon the assumption that this relation is linear,

you can treat this as being given two ordered pairs in the relation, namely

(40,50 and (80,60), where the first variable is c (chirps) and the second variable is t (time)

so we are looking for t = mc + b (to resemble the familiar y = mx + b linear equation)

slope = (60-50)/(80-40) = 10/40 = 1/4

using m= 1/4 and the point (40,50)

50 = (1/4)(40) + b

b = 40

so you have t = 1/4c + 40

when c = 95

t = (1/4)(95) + 40

= 63.75 degrees

OR

draw a straight line, plot the points

(40,50), (80,60) and (95,t)

then (t-60).(95-80) = (60-5)/(80-40)

(t-60)/15 = 1/4

4t - 240 = 15

t = 63.75

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