# Spinnerz

• Nov 29th 2008, 01:32 AM
kurdupel
Spin
Need help with this problem. Thanks.

Problem 1:

4 numbers on the spinner are equally likely. They are also all different.

Smallest sum you can get in two spins of this spinner is 2; largest is 14.

impossible to get odd number if you spin this spinner two times and add results.

If you spin this spinner twice, and add the two numbers, you are likely to get 10 as 6.

impossible to get an even number on this spinner if you spin three times and add.

most likely sum of two spins of this spinner is eight.

Draw spinner from these hints.

Problem 2:

spinner has 3 sections. While no 2 are the same size, one of the sections is half the size of another.

You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.

largest number you get in 2 spins of this spinner is one.

If you spin the spinner twice and add, you get a sum of one about a quarter of the time.

sum of 3 numbers on the spinner is one.

If you spun the spinner 100 times and added up all the numbers, you'd probably get somewhere near 40.

Draw spinner from these hints
• Nov 29th 2008, 01:52 AM
mr fantastic
Quote:

Originally Posted by kurdupel
Need help with this problem. Thanks.

Problem 1:

4 numbers on the spinner are equally likely. They are also all different.

Mr F says: What does this tell you about the size of the four sections.

Smallest sum you can get in two spins of this spinner is 2; Mr F says: What number does that tell you is on the spinner ....? (Hint: If you get it twice the sum is 2 .....)

largest is 14. Mr F says: What number does that tell you is on the spinner ....? (Hint: If you get it twice the sum is 14 .....)

So now you know two of the numbers.

impossible to get odd number if you spin this spinner two times and add results.

Mr F says: This should tell you that the remaining two numbers are odd.

If you spin this spinner twice, and add the two numbers, you are likely to get 10

Mr F says: What number does this tell you is on the spinner given the two numbers you so far know are on the spinner?

as 6.

Mr F says: What number does this tell you is on the spinner given the two numbers you so far know are on the spinner?

impossible to get an even number on this spinner if you spin three times and add.

Mr F says: Confirm this statement with the four numbers you now have.

most likely sum of two spins of this spinner is eight.

Mr F says: Confirm this statement with the four numbers you now have.

DRAW SPINNER.

Problem 2:

spinner has 3 sections. While no 2 are the same size, one of the sections is half the size of another.

You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.

Mr F says: Possible totals for two spins are 2/3, 1 and 5/6. Given one of the numbers that you know is on the spinner (see below), what does this tell you the other two numbers must be?

largest number you get in 2 spins of this spinner is one.

Mr F says: What number does that tell you is on the spinner ....? (Hint: If you get it twice the sum is 1 .....)

If you spin the spinner twice and add, you get a sum of one about a quarter of the time.

sum of 3 numbers on the spinner is one.

If you spun the spinner 100 times and added up all the numbers, you'd probably get somewhere near 40.

DRAW SPINNER.

For Problem 2 you should now know the three numbers that are on the spinner. The rest of the information is telling you how relatively large the section is that each of these numbers is in.
• Nov 29th 2008, 02:09 AM
kurdupel
Thanks...how do I draw it though?
• Dec 10th 2008, 09:30 PM
kurdupel
- Hi,

After looking at this question I think you;ve got it wrong. If you read the question for 3a it says The four numbers on the spinner are equally likely. They are also all different.

So therefore they spinner can't have 7,7,1 and 1 configuration.

I thinkt the spinner should look something like this....

http://img392.imageshack.us/img392/1892/circleqw8.png

Would you agree?
• Dec 10th 2008, 11:27 PM
mr fantastic
Quote:

Originally Posted by kurdupel
Hi,

After looking at this question I think you;ve got it wrong. If you read the question for 3a it says The four numbers on the spinner are equally likely. They are also all different.

So therefore they spinner can't have 7,7,1 and 1 configuration.

I thinkt the spinner should look something like this....

http://img392.imageshack.us/img392/1892/circleqw8.png

Would you agree?

Your spinner fails to address most of the criteria ..... eg. Two spins of your proposed spinner cannot give 14 .....

Read my post again - carefully. I never said or suggested that there were two 1's and two 7's .......

Note: If there's one 1, it's possible to spin twice and get 1 twice .....
• Dec 11th 2008, 02:06 AM
kurdupel
- Thanks...

I looked at all the requirements and this one seems to fit the best for 3a..don't know if its correct though...

http://img139.imageshack.us/img139/9808/23693019jv2.png
• Dec 11th 2008, 02:36 AM
kurdupel
- For 3b I worked it this way (still I'm not sure)...never been confident with my mathematics.

-largest number you get in 2 spins of this spinner is one.
Therefore If I get it twice the sum is 1, if I get it once it is 0.5
-If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
Therefore If I get it twice the sum is one a quarter of the time (0.25), therefore once is half that…0.125
- sum of 3 numbers on the spinner is one.
To get the rest I simply added 0.125 and 0.5 to get 0.625 and subtracted this from 1 to get 0.375.

TO DRAW THE EQUIVALENT SPINNER I MULTIPLIED THE VALUES….0.5, 0.125 and 0.375 by 360 degrees (spinner).
0.5 × 360 = 180 degrees.
0.125 × 360 = 45 degrees.
0.375 × 360 = 135 degrees,

http://img377.imageshack.us/img377/9541/10741470kr9.png
• Jan 5th 2009, 10:04 PM
kurdupel
SpinnerProblem
- Hi I am having trouble with a spinner problem..I think I've worked it out I just need a Math whiz to see that I am right (or wrong)?

Thanks..

Problem 1 Spinner

http://img139.imageshack.us/img139/9808/23693019jv2.png

Problem 2 Spinner

-largest number you get in 2 spins of this spinner is one.
Therefore If I get it twice the sum is 1, if I get it once it is 0.5
-If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
Therefore If I get it twice the sum is one a quarter of the time (0.25), therefore once is half that…0.125
- sum of 3 numbers on the spinner is one.
To get the rest I simply added 0.125 and 0.5 to get 0.625 and subtracted this from 1 to get 0.375.

TO DRAW THE EQUIVALENT SPINNER I MULTIPLIED THE VALUES….0.5, 0.125 and 0.375 by 360 degrees (spinner).
0.5 × 360 = 180 degrees.
0.125 × 360 = 45 degrees.
0.375 × 360 = 135 degrees,

http://img377.imageshack.us/img377/9541/10741470kr9.png

Thanks.
• Jan 9th 2009, 03:54 AM
mr fantastic
I would formally and publicly like to express my appreciation of the professionalism and respect shown by Moderator Stapel (Liz) at another Forum that the OP copied and pasted posts from this thread into.

And to Constatine11, who's fine hand I detect in all this.

Thankyou both.

Since the OP deleted all his/her posts in this thread, and in light of other information, I have drawn the conclusion that the OP was attempting academic fraud. This thread is closed.