course Mth 163
Assignment 2 Exercises1. From the previous assignment I determined that the parameters were a=.00534 b= =1.33 and c=63.496
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
To predict the depth when clock time is 46
y=.00534(46^2)+(-1.33)46+63.496
y=.00534(2116)+(-1.33)46+63.496
y=11.299 + -61.18+63.496
y=13.615 cm
When the clock time is 46 seconds the depth is 13.615cm or (36,13.615)
To determine the clock time when the depth reaches 14cm.
14=.00534(t^2)-1.33t+63.496
14-14=.00534t^2-1.33+630496-14
0=.00534t^2-1.33+49.496
t=[1.33+-sqrt(-1.33^2-4(.00534)(49.496)]/2(.00534)
t=[1.33+-sqrt(1.769-1.0592)]/.01068
t=[1.33+-sqrt(.842)]/.01068
(1.33+.842)/.01068=203.37
(1.33-.842)/.01068=45.93
So t=45.93
When the depth reaches 14 cm the clock time will be 45.93 secs.
I believe that the model fits the data well, and the results fit nicely with the predictions on the table.
2. DATA SET 1
I used coordinates (0,1) (10, 1.790569) and (20, 2.118034) for the following model procedure.
Solving for A
1=a(0^2)+b0+c
1.790569=a10^2+b10+c
2.118034=a20^2+b20+c
a+b+c=1
a100+b10+c=1.790569
a400+b20+c=2.118034
a100+b10=1.790569
a0+b0=1
a100+b10=.790569
a400+b20=2.118034
a100+b10=1.79056
a300+b10=.32747
10(a100+b10)=(.790569)10 = 1000a=7.90569
-10(a300+b10)=(.32747)-10= -3000a=-3.2747
-2000a=4.63099
a=-.00232
Solving for B
-.00232(100)+b10=1.790569
-.232+b10=1.790569
b10=2.0226
b=.20226
Solving for C
-.00232(0)+.20226 (0) +c=1
c=1
I have now determined that the parameters are a=-.00232 b=.20226 and c=1
To determine from the model the percent of classes reviewed to schieve grade of 3.0 and 4.0 I will need to solve for t
3=-.00232t^2+.20226t+1
To change it to standard form will subtract 3 from each side.
0=-.00232t^2+.20226t-2
Solving for t
t=[-.20226+-sqrt(.0409-4(-.00232)(-2)]/-.00464
t=(-.20226+.0476)/-.00464=33.332
t=(-.20226-.0476)/-.00464=53.849
t=53.849 will fit best with the model
use the same equation for 4 and the result was t=68.219
Students reviewing 53.849 percent of their assignments could receive a 3.0 GPA and those who reviewed 68.219 percent of their assignment could receive a 4.0 GPA
y=-.00232(80^2)+.20226(80)+1
y=-.00232(6400)+16.1808+1
y=2.3328 grade average
A student reviewing 80% of the assignments could receive a 2.3328 GPA
I did not think that this data set fit the model well. I determined they could only receive a 2.3328 GPA from reviewing 80% but that they could get a 4.0 from only 68.219
so the results were not consistant.
DATA SET 2
From the model we can determine what illumination would be expected at 1.6 Earth distances from the sun.
I used data points (1, 935.1395) (2, 264.4411) (3, 105.1209)
Solve for a
a1^2+b1+c=935.1395
a2^2+b2+6=264.4411
a3^2+b3+6=105.1209
4a+2b=264.4411
1a+1b=935.1395
3a+1b=-670.6984
9a+3b=105.1209
4a+2b=264.4411
5a+1b=-159.3202
1(3a+1b)=-670.6984(1)
-1(5a+1b)=-159.3202(-1)
3a=-670.6984
-5a=159.3202
a=255.6891
Solve for b
255.6891(4)+b2=264.4411
1022.7564+b2=264.4411
b2=-758.3153
b=379.15765
Solve for c
(255.6891)1^2+c=935.139
255.6891+379.15765+c=935.1395
c=300.29275
The parameters for the model are a=255.6891 b=379.15765 and c=300.29275
To determin the illumination I must solve for t
935.1395=255.6891t^2+379.15765t+300.29275
0=255.6891t^2+379.15765t-634.84675
t=[-379.15765+-sprt(143760.5236-649293.4766)/511.3782
t=-379.15765+-sqrt(143110.5878)]/511.3782
t=-379.15765+378.299/511.3782=-.00168
t=-379.15765-378.299/511.3782=-1.481
t=-1.481 fit the model the best
y=255(2.0107)-267.389+300.29275
y=545.73225
The illumination at 1.6 Earth distance would be 545.73225.
The range of 25 to 100 Watts per square meter would fall in the 3-6 AV range.
This set of data fit the model very well.
"
Your work here is excellent. Note, however, that it's probably easier for you to use the Query program, which includes problem statements and asks specific questions about selected problems, than to submit the problems in this format. Only the Query is required. You aren't required to submit problems in this format.
However if you prefer this format, however, it's fine with me.
a