linear transformation model help

- jessicawade

linear transformation model help

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- jessicawade

Number of months Number of fish
0 8
1 39
2 195
3 960
4 4,738
5 23,375

- jessicawade

log y hat=0.9013x+0.6935

- jessicawade

Use the linear transformation model to predict the number of fish in 12 months.

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## More answers

- jessicawade

i got 5,685,367

- jessicawade

@atreyu6s3x6

- anonymous

haha sorry im bad at prob and stats

- mathmate

The model is way off even at 5 months. Can you double-check the model?

- anonymous

i think @mathmate has it haha

- jessicawade

hmmm thats what i got

- jessicawade

(log y hat=0.9013•logx+ 0.6935 ) this the other model thats is the closest to the one i have

- jessicawade

but im pretty sure my log is correct

- jessicawade

I have this question and the other one @atreyu6s3x6 tried helping me with. i can show you the entire part of the lesson if you need it.

- jessicawade

Scientists are studying the population of a particular type of fish. The table below shows the data gathered over a five–month time period. Use the data to answer questions 5–9.
Number of months Number of fish
0 8
1 39
2 195
3 960
4 4,738
5 23,375
5. What does the scatterplot of the data show? (1 point)
• a strong positive linear relationship
• a strong negative linear relationship
• a curve that represents exponential growth *
• a curve that represents exponential decay
6. Complete an exponential transformation on the y-values. What is the new value of y when x = 5?
(1 point)
• 4.3688
• 3.6756 *
• 0.6990
• 3.3757
7. Find the linear transformation model. (1 point)
• logy hat=o.6935•logx+ 0.9013
• log y hat=0.9013x+0.6935*
• log y hat=0.6935x+ 0.9013
• log y hat=0.9013•logx+ 0.6935
8.
Use the linear transformation model to predict the number of fish in 12 months. (2 points)

- jessicawade

i put a * next to my answers

- mathmate

It is much clearer when you post the complete original post.
If you post your answer as part of the question, it will make the question inconsistent.
First, do you think it is a linear or exponential relationship?

- jessicawade

well i think its linear because when the x values increase so do the y values

- jessicawade

JUST KIdding

- jessicawade

its an exponential growth lol

- mathmate

exactly!
What have you learned about transformation?

- jessicawade

not sure, i have taken notes but i lost the notebook earlier yesterday

- mathmate

Do you have a textbook?

- jessicawade

no im on online school

- mathmate

You cannot go back to the lessons?

- jessicawade

I can but they wont explain everything once i pass the lesson just bits

- jessicawade

so i got log y hat=0.9013x+0.6935 as the transformation model out of the answers, but im not sure how to find the number of fish after 12 months, which is confusing me lol

- mathmate

ok, are you looking for the answer or are you looking to understand?

- jessicawade

understand please

- mathmate

We'll rewind to the beginning, ok?

- jessicawade

ok

- mathmate

Typically, a linear model has the form
y=ax+b
but that's not our case.

- jessicawade

ok

- mathmate

Similarly, an exponential growth model has the form
\(y=ax^{bx}\)
where a and b are to be found.

- mathmate

so far so good?

- jessicawade

yes

- mathmate

However, the parameters a and b are hard to calculate directly from the exponential model. Since we (including you) already know how to model a straight line, we will transform the exponential to a straight line. Then we'd find the parameters as though it is a straight line.|dw:1447097415267:dw|
That's where transformation comes in.

- mathmate

@jessicawade
are you still there?

- jessicawade

yeah my computer wasnt laoding

- jessicawade

ohhhh ok :)

- mathmate

The way the transformation works is you would take log (to base 10 in your case) on both sides.
Can you do that for me?

- mathmate

Take log on both sides of
\(y=ax^{bx}\)

- jessicawade

let me try im not very good at math

- jessicawade

so on both sides on the y and the ax?

- jessicawade

i dont get it xD

- jessicawade

@mathmate

- mathmate

\(y=ax^{bx}\) actually should read
\(y=a(10)^{bx}\)........ if we take log 10 eventually
We'll take log on both sides, so
\(log(y)=log(ax^{bx})=log(a)+log(10^{bx})=log(a)+bx~log(10)=bx+log(a)\)
This is done by the laws of logarithm (which you'll need to brush up for this course)
Put it simply,
\(log(y)=bx+log(a)\).........where a and b are constants to be found for the given data set.

- mathmate

So we just finished the transformation part. Except for the laws of logarithm, are you following with the concept?

- jessicawade

yeah so far i think haha

- mathmate

It turns out that the constant "a" is the initial value, or the y-intercept.What is the y-intercept in our problem?

- jessicawade

is that the same as log y?

- mathmate

|dw:1447098799871:dw|
"a", the y-intercept is the value of y when x=0.
In our case, a=8 becase the number of fish is 8 at month 0.

- jessicawade

ohhhhhhh!!!

- jessicawade

so a and y are basically the same thing

- mathmate

So what is now our equation?
from
\(log(y)=bx+log(a)\)
becomes?

- jessicawade

so we just have to find the number of fish in 12 months, that doesnt seem hard but i dont know how to find it lol

- mathmate

You need to know (or confirm) that you're using the right model before you can find the number of fish at 12 months, right?

- mathmate

Can you find what is log(8)?

- jessicawade

correct, so thats what were trying to do right? find the correct model, or try to see if mine is correct

- mathmate

We'll know if we're finding one or confirming one after the next step.
Can you find the value of log(a)=log(8)?

- jessicawade

hold on sec

- jessicawade

no ium stuck

- jessicawade

because isnt a=8

- jessicawade

sorry im being complicated.. lol

- mathmate

a=8 is right, but we need log(8) in the model.
If you don't have a calculator, google log(8).

- jessicawade

ok

- jessicawade

its 0.90308998699

- mathmate

Good, so let's call it 0.9031.
So the model now becomes:
log(y)=bx+0.9031
Can you now go back to the list of 4 models and check which one is appropriate?

- jessicawade

yeah none has y=0.931

- jessicawade

i mean 0.9031

- mathmate

Well, remember that y_hat is just approximate estimation of y, so 0.9031 and 0.9013 are close enough for a model.
Can you decide which one fits the bill? There is only one that fits (approximately) the model:
log(y)=bx+0.9031
where b is a constant to be found (but will be provided by the model we choose).

- jessicawade

hmmm

- jessicawade

• log y hat=0.9013x+0.6935*
• log y hat=0.9013•logx+ 0.6935

- jessicawade

both the y= 0.913 but im not sure

- mathmate

• logy hat=o.6935•logx+ 0.9013
• log y hat=0.9013x+0.6935*
• log y hat=0.6935x+ 0.9013
• log y hat=0.9013•logx+ 0.6935
Here are the four choices to match
log(y_hat)=bx+0.9031
You need to match the constant (at the end).
That leaves you with two choices, which two?

- mathmate

A constant is a number which is not multiplied with a variable (x, or y)

- jessicawade

ohhh

- jessicawade

a and c

- jessicawade

so logy hat=o.6935•logx+ 0.9013 or log y hat=0.6935x+ 0.9013

- mathmate

Good, now look carefully, which model out of the first and third resembles our model closely?

- jessicawade

log y hat=0.6935x+ 0.9013

- jessicawade

so 3

- mathmate

Exactly, that fits what you chose earlier, however you chose it.
So we've got a model, right?

- jessicawade

yeah now we need to find the number of fish in 12 months

- mathmate

We're almost there, just one more step.

- jessicawade

ok!

- mathmate

we now have the model
log y hat=0.6935x+ 0.9013
Using the inverse transformation (i.e. transform back), we need to use the law of logarithms again to get:
y_hat = 10^(0.6935x+0.9013)
But whenever we work with a model, we need to check it out.
We'll check for x=0,
y_hat(0)=10^(0.6935x+0.9013)=10^(0.6935*0+0.9013)=7.967, close enough to 8 that we're expecting.
We need to check another one, say x=5 months, can you do that?

- jessicawade

let me try

- jessicawade

plug in 5 for x?

- mathmate

exactly!

- jessicawade

i cant calculate it

- jessicawade

its not giving me an answer lol

- mathmate

type this in google
10^(0.6935*5+0.9013)

- mathmate

lol you still need to learn to use your calculator!

- jessicawade

ohhh lol *facepalm*

- mathmate

what, it didn't work?

- jessicawade

i got 23377.6041227

- mathmate

is this close to what was expected for 5 days?

- jessicawade

the first part yes lol

- mathmate

* months, not days lol
Now you're ready for the plunge..... calculate in the same way what you expect to see in 12 months.

- jessicawade

so i got 1,672,245,362
o_o

- mathmate

Yes, it's a big number, and of course a very optimistic estimate. There will not be enough food for all the fish.
But....mathematically you're right! yay!

- jessicawade

yay!!! lol

- jessicawade

ok i hhave 1 more

- mathmate

One last word!
Please
1. learn to use your calculator, using powers and log
2. learn the laws of exponents and logarithms.

- mathmate

I have to go,but I can give you hints to get started, how's that?

- jessicawade

ok!

- mathmate

I suggest you start a new post, so others may be able to help as well.

- jessicawade

i tagged you in one, i have 2 people helping me but they are confusing me lol

- mathmate

ok, I'll see what I can do!

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