Here's the question you clicked on:
julliemyers
This may seem stupid but can anyone help?? I have to divide 10 by 5 minutes and 54 seconds????
\(5 \min 54 \sec = 5 \dfrac{54}{60} = 5.9\)
Im doing a Algebra2 experiment and i have to divide my average time by 10 Here are my times 5 min 54 sec 4min 55sec 3min 30sec so when i divide these all by 10 do i put the time in decimal form? Like 5min 54 sec= 5.54??
Essentially, \(5 \min 54 \sec = {5 } + \dfrac{54}{60}\) as we're expressing the seconds as a fraction.
Now, you can simplify \(\dfrac{54}{60}\).
so, 5.9 divide by 10? and i get .59? and that would be my awsner for that one
BTW, it'd be dividing 5 min 54 sec by 10, not the vice-versa.
Yes, that'd be your answer :) @julliemyers
oh ok then so just to make sure i get the rest correct let me do them and will u check them?
hmm it says to divide the average time by 10.. so that might not be the answer.. first you have to convert all your times into decimal
((10/5) minute) + (54 seconds) = 2.9 minutes
Didn't see that question below.
no i had 3 trials and i have to divide 10 by indivaual times
serach on google divide 10 by 5 minutes and 54 seconds
Okay... so can you divide the other times by 10 after converting them and take the average of all figures you get?
hang on ill show u my table
5 minutes 10 seconds divide it by 5 = 1 minute 2 seconds.
Oh, so that is the question.
this is the directions . For each trial, use a stopwatch to record the number of seconds it takes for the pendulum to complete 10 full swings. Record each time in the i rst column of the table provided on the next page. Next, i nd and record the average of the three times you listed. Finally, divide the average time by 10 to determine the period of the pendulum. Repeat the procedure using two coins, then using three coins, recording the data in the second and third columns, respectively. Does it appear that the weight of the bob af ected the period of the pendulum? What factors other than the weight might af ect the period of the pendulum?
So you have to find the average period. It'd be 5 min 54 sec by 10.
yes each collum had their own average time
Then I'm afraid you'd have to complete this experiment by yourself.
Did you do this experiment?
hmm ok i have a question.. something don't seem right. in the trial 1 with 1 coin the time u have wrote is for 10 swings?
yes it was in the file i attatched i just needed help on how i would divide time by 10
But that wouldn't be the time for 10 swings. Though there's a cheat: You can divide it by 3. ;)
add all noted times with seconds as fraction and then divide by 10. here is your answer :)
@julliemyers wel i think the average time u have found for 10 swings in that file is wrong.. u need to check that again.
You can add 3 noted times and divide by 3 -- no one would realize :D
@nubeer: Seems like julliemyers has summed up the three noted times in the above rows.
yup she did that.. she just forgot to divide by 3
@nubeer it doesnt say i need to divide by 3 tho?
i know it don't say and it will never say.. the Word Average itself says it.. Average = Total time/ no. of times experiment performed.. means you would add up the time of all the 3 trials and then divide it by 3.. IF INCASE.. there was a 4th trial then you would add all the 4 times and divide by 4... you have to do this where ever the average word comes up
ohh Yea i remeber that. So for ex 5 min 54sec would be 5.54/3 =1.846 then i devide that by 10?? =0.1846????
ok i think i have got it lol -.-
@ParthKohli No to what? Im wrong?????
lol that was 5.9 .. the total time.
It's not \(5.54\); it's \(5.9\)
And you don't have to further divide by 10.
lol she have to divide by 10 to find the time period.. @ParthKohli
Im so confused (-.-)
First add all the time , trial 1,2,3 that would be 5.9 and divide it by 3.. the value you will get you will put that against this row "Average time complete 10 full swings"
Now.. the value in this row "Average time complete 10 full swings" you would divide it by 10 and put them in this row "Period of pendulum "
@nubeer: My medal increased your SS :P
lol does it ? really? hahah i dont think so.. lol i never work on medals but still thank you...
Im spose to have 3 periods of pendulum times, I dont think i add up all my averages. i think im spose to have 3
hmmm .. :( not getting you.. ok look the file you have posted and see the row average time for 10 swings what value u have in there?
yes those r the values added of the coin 1 coin 2 and coin 3
ok.. now 5 min 54 secs convert in decimal you will get 5.9 right? you did that before now divide 5.9 by 3... becasue that total time is for 3 trials.. getting it?
i get 1.966666666 = 1.97
so 1.97 would go in the period of pendulum collum???
ok.. good.. now this is average time for 10 swings.. ok? now for the time period of pendulum.. this means the time period for just 1 swing so u divide 1.97 by 10
Should i have 5.9, 4.9, 3.5In the average collum? Instead of the minutes and seconds or should i leave it how it is
hmm i told u these are not average.. 5.9/3 would be the aveage so 1.97 mins is the average of 10 swings.
ok thats what i was wondering lol i feel stupid but i understand it now lol
ok so i fixed it take a look
yup looks ok now.. this all are in mins. :)