Maths question in exam todau
- Thread starter billybob89
- Start date
E = at + b is just like the standard linear equation y = mx + c
(a/m is the gradient, b/c is the y intercept)
There are two ways to do this but the easiest one here is to first find the gradient using the formula:
Method 1
m(the gradient, a in this case) = y-y1/x-x1
Taking any two sets of data (using two as far apart as possible gets the most accurate result, I used the first and the last) from what you've been given you get:
(30.2 - 6)
----------- = 0.03025
(1000 - 200)
That's your gradient (or a), so, right now you have
E = 0.03025t + b
You have E and t values so substitute any one of them in to get b
E = 14.8 t = 500
b = 14.8 - (0.03025 x 500)
= -0.325
To check (use another set of data, I chose t = 900 E should = 26.9):
E = 0.03025 x 900 - 0.325
E = 26.9 (value in data is actually 26.8)
This is not as accurate as using simultaneous equations though (The other method to solve this) which yields exact answers where a = 0.03 and b = 0 shown here:
Method 2
Two unknown variables (a and b) means making two equations using the data you've been given
I used the first and the third:
(t = 200 E = 6, t = 400 E - 12)
6 = 200a + b and
12 = 400a + b
You can eliminate b by rearranging:
b = 6 - 200a and (1)
b = 12 - 400a (2)
so 6 - 200a = 12 - 400a
rearrange (take -400a to left and +6 to the right) to get:
200a = 6
a = 6/200
Substitute that back into one of the equation, say (1)
b = 6 - 200(6/200)
b = 6-6 thus b = 0
You can check this by substituting these values back into the original equation:
E = 0.03(t) + 0 (note 6/200 = 0.03)
Take any value from your data say, t = 700, you get E = 4200/200 + 0
E = 21.0 as shown in your data table.
Hope this makes sense, It's easier than it looks, I just put in all the steps so you could understand.
Simultaneous equations was my thought also, no time to work through it though
Simultaneous equations was my thought also, no time to work through it though
Simultaneous equations will not work, a casual glance at the numbers should tell you that they don't exactly follow at+b, instead they're distributed around the line you're to find.
It's a linear regression question. For a and b you can draw it by hand and measure the gradient for a and the y-intercept for b. For the "test if it is so" part you would have to work out the correlation co-efficient using a formula I can't be bothered to type here....
......or whack it into stats software like I just did and get
E= 0.03005T - 0.0078 with an R2 of 100% and p vale of 0.000
......or whack it into stats software like I just did and get
E= 0.03005T - 0.0078 with an R2 of 100% and p vale of 0.000
Sim Equations gives me
a=0.03
b=0
as 6=200a+b and 21=700a+b
although the trend is not linear and the equation does not seem to hold.
300*0.03+0 = 9, which is not the same as test data.
I suppose that would be the answer to the question.
a=0.03
b=0
as 6=200a+b and 21=700a+b
although the trend is not linear and the equation does not seem to hold.
300*0.03+0 = 9, which is not the same as test data.
I suppose that would be the answer to the question.