Question

please I need help I have no idea how to do this and would appreciate an explanation not just the answer


Archers need to use arrows that do not bend easily. Th e table shows how the weight of
an arrow affects its spine, or the distance the center of the arrow bends when a certain
constant weight is attached. Graph the data in the table to find a linear and a quadratic
model for the data. Use the regression feature on your calculator to find each model.
Which model is a better fit? Explain.

weight (in grams)140,150,170,175,205
weight (in inches)1.4, 1.25, .93, .78, .43

Answer 1

do you have a calculator? ti83 or such?

Answer 2

yes I do

Answer 3

you are going to make 2 lists ... input the data given. does that make sense?

Answer 4

just on the main screen or in the y= section?

Answer 5

stat, edit ; should get you into the lists

Answer 6

ok got that

Answer 7

enter the data into the lists ..

L1 L2
140 1.4
150 1.25
170 .93
etc ...

Answer 8

ok done

Answer 9

2nd mode ... to quit back to main screen

then, stat, calc, pick a regression

Answer 10

after you pick a regression, you tell it how to view the lists:

LinReg(ax+b) L1, L2

or L2, L1 if you want it oriented the other way

Answer 11

ok

Answer 12

it should give you values for a, b on the linear model; and a,b,c if you picked the quadreg

Answer 13

LinReg(ax+b) L1, L2


a = -.0152...
b= 3.5125...

Answer 14

??? it shows "LinReg(ax+b) L1,L2

Answer 15

after you input the L1,L2 .. you hit enter

Answer 16

ERR: DATA TYPE

Answer 17

stat
calc
linreg(ax+b)
2nd, 1
2nd, 2
enter

linreg(ax+b)
y = ax+b
a= ....
b= ....

Answer 18

if you are just inputing alpha L ... i dont think it knows that you are looking for List1

Answer 19

2nd, 1 for L1

Answer 20

ohhhhhhhhh lol gotcha

Answer 21

quadreg
a=5.6706282E-5
b=-.0347792869
c=5.171748727


so after I have that and the linrereg what do I do?

Answer 22

well, whose give you the coefficients for the quad or line equations

id enter them to form the equations

Answer 23

assuming we want an x axis of L1, and a y axis of L2 of course

Answer 24

if we do an L2, L1 ordered pair ...

a = 16.1966
b = -94.89
c = 242.12

to give us a quadratic as:

ax^2 + bx + c

Answer 25

so I would put those values into the equation and solve for X?

Answer 26

no need to solve for x, x is the input in terms of L2 values for the one I ran

it lets us approximate/predict what the value of an unknown L2 value should be

Answer 27

16.1966x^2+-94.89x+242.12?

Answer 28

the equation:

y = ax^2 + bx + c ; for some x=L2

yes

Answer 29

determining which model is a best fit might depend on the original plotting of the points .... do you ahve excel?

Answer 30

excel?

Answer 31

ill assume you dont :)

Answer 32

ok so is that the original points you mentioned

Answer 33

thats the original points they gave us yes

Answer 34

im not too sure which one is a better fit tho ... might be something to do with a correlation coefficient that i just do not have that much experience with

Answer 35

do you mind helping me with something else ?

Answer 36

i spose that depends on what the something else is ....

Answer 37

An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path. the top of the target is 1.5 meters high.


the target is 20 cm from center to top

Answer 38

you have 3 points ... wrap a quadratic to them ...

Answer 39

ive done this question on here a few times

Answer 40

if we do the calculator again:

L1 L2
0 shoulder
18 target aim
45 0

Answer 41

this would help define the parabolic path, and from there we can determine its highest point

Answer 42

theres prolly a pic of a target with rings of 4,4,4,4,8

Answer 43

haha yes there is the picture

Answer 44

the solution results are ugly to say the least ... workable, but awful to look at

Answer 45

0, 1.39
18, 1.5-.24
45, 0

Answer 46

ok?