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Question 1194267: In a laboratory experiment, it is observed that when the temperature is reduced 𝑇 (degrees Celsius) of a rabbit, its heart rate (beats per minute) decreases. Under laboratory conditions, a rabbit at a temperature of 37°𝐶 has a heart rate of 220 and at a temperature of 32°𝐶 its heart rate heart rate decreases to 150. The function 𝑅 is heart rate and is linearly related to the temperature 𝑇. a) Find the functional relationship between 𝑅 and 𝑇 and its respective graph. b) Find the variation of 𝑅 if 𝑇 = [26, 38].
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Given info:
Temperature = 37 degrees C ---> heartrate = 220 bpm (beats per minute)
Temperature = 32 degrees C ---> heartrate = 150 bpm
R = heart rate in bpm
T = temperature in degrees C
Since R is linearly related to T, this means a straight line goes through the two points (37, 220) and (32,150)
Each point is of the form (T, R) with T being the input independent variable and R is the output dependent variable.
Let
x = T
y = R
So each point is now of the form (x,y)
Let's find the equation y = mx+b
We'll first need the slope m
m = slope
m = rise/run
m = (change in y)/(change in x)
m = (y2-y1)/(x2-x1)
m = (150-220)/(32-37)
m = -70/(-5)
m = 14
The slope is 14 to indicate each time the temperature increases by 1 degree, the heart rate goes up by 14 bpm.
In other words, it increases by 14 bpm per degree.
Now find the y intercept
We'll use the slope we just found and let's say the point (x,y) = (37,220)
y = mx+b
y = 14x+b
220 = 14*37+b
220 = 518+b
220-518 = b
-298 = b
b = -298
This is the bpm when the temperature is 0 degrees C.
Of course it's not realistic because we can't have a negative bpm. However, it does help set up the line. We'll just need to be careful on how to set up the domain.
Since m = 14 is the slope and b = -298 is the y intercept, we go from y = mx+b to y = 14x-298
Then that leads to R = 14T - 298
Let's see what happens when we plug in T = 37
R = 14T - 298
R = 14(37) - 298
R = 518 - 298
R = 220
As expected, a temperature of 37 degrees C leads to a heart rate of 220 bpm.
Now try T = 32
R = 14T - 298
R = 14(32) - 298
R = 448 - 298
R = 150
This works out as well. A temperature of 32 degrees C leads to a heart rate of 150 bpm.
This paragraph and the previous one helps fully confirm we have the correct linear function to connect R and T together.
Answer: R = 14T - 298
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Part (b)
I'm not sure what you mean here. I think something got lost in translation.
Though if you wanted to know the range of bpm values for the given temperatures, then,
Plug in T = 26
R(T) = 14T - 298
R(26) = 14(26) - 298
R(26) = 66
Do the same for T = 38
R(T) = 14T - 298
R(38) = 14(38) - 298
R(38) = 234
Find the difference:
R(38) - R(26) = 234 - 66 = 168
The range of bpm values is 168 when considering the temperature interval [26, 38] aka 
Once again, I'm not sure what you're asking for this part (b).
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