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- bobbym
- 2013-01-25 05:02:35
Hi Ella123;
You must mean 20 km/h for 2hrs and 60km/h for 30 minutes.
2 x 20 km = 40 km
( 1 / 2) 60 km = 30 km
You travelled 70 km in 2 and 1 / 2 hrs. So you divide the 70 by ( 2 + 1 / 2 ) and get 28 km / hr. That is your average speed.
- Ella123
- 2013-01-25 04:17:37
For my maths homework I need to find the average speed. I tried to add up and divide but the answer did not look right . Here are the speeds:
20 km/h and 60km/h.
I added up 20 and 60 and got 80 and then divided by 2. I them had 40km/h. As the question said that they travelled at 20 for 2 hours and 20 for only 30 minuets I do not see how 40km/h could be the answer.
Help me please!
- REMA
- 2013-01-14 05:45:53
bob...thanks for the explanation. Clear and concise
- bob bundy
- 2013-01-14 05:28:18
hi REMA
Welcome to the forum.
This is called a weighted average. By applying a 'weight' to each type of medal you can take account of it being 'better' to get a higher weighted medal.
So if you wanted to 'score' countries in the Olympics, for example, your system of G = 3, S = 2 and B = 1 would make more sense.
But the purpose is to get the average placement where 1st is best and then 2nd and then 3rd. So you want the 'value' of gold to be 1.
Take an extreme example. If a player got only golds, say 20, then the calculation would be (20 x 1)/20 = 1, so the average placement would be 1, ie. the best.
If the player got 19 golds and 1 silver the calculation would be (19 x 1 + 1 x 2)/20 = 21/20 = 1.05 so the average placement has dropped a bit.
But you do need to be told the weightings.
And the system has a serious flaw. If a player gets 1 gold and 19 below bronzes the average is (1 x1 + 10 x 0)/20= 0.05 which is even better than the 20 gold player. Hhmmmm!
So you are right to be confused. So am I. 
Bob
- REMA
- 2013-01-14 04:24:41
Hello Can you please help with this:
A person wins 27 gold medals , 17 silver and 2 bronze over a period of years. (total 46 medals)
The task is to find the average placement.
It was solved by doingthis sum: 27(gold)*1 = 27 , 17(silver)* 2 = 34, 2(bronze) * 3 = 6
add 27+34+6 = 67 then divide 67 / 46 medals giving an average placement over the persons 46 races of 1.456 or
on average they came in 1.5 st place (rounded up)
What we can't get is how do you/why do you assign the values to the medals i.e Gold =1 (i-e first place) , Silver =2 (2nd place) and bronze = 3 (third place)
WE were not told that the medals carried seperate values...so we could've decided that gold = 3 (i.e highest points or even 100) and bronze = 1 (i.e lowest value, or even 0)
Can you explain why it was ok to "assume" gold = 1 etc etc...many thanks (it's my 13 yrs olds problem.)
- mathgogocart
- 2012-12-10 08:40:03
Good Luck.
- bob bundy
- 2012-12-10 08:34:57
hi ihtemthsbtneedtoasksmthng
Sorry, in my rush I forgot to say
Welcome to the forum.
Good luck with the exams.
Bob
- ihtemthsbtneedtoasksmthng
- 2012-12-10 07:26:06
Thank you so much! I'll try and remember that for the exam tomorrow!
- bob bundy
- 2012-12-10 06:53:56
Yes that's right. Notice the answer lies between 10 cm and 20 cm, but nearer to 10 because there were more months at that rainfall.
So, can you do B ?
Bob
- ihtemthsbtneedtoasksmthng
- 2012-12-10 06:46:32
so would A be 13.3?
- bob bundy
- 2012-12-10 06:29:07
Ok. think about the information. You have 12 months data altogether; 8 then 4 months for A and 10 then 2 months for B.
You need to work out how much rain fell in A altogether over the year.
For 8 months it was 10 cm every month. How much is that altogether?
Then 4 months at 20 cm every month.
Then add them together and divide by 12 for the average for A.
Bob
- ihtemthsbtneedtoasksmthng
- 2012-12-10 06:21:43
Which of the places has more rainfall on average over the whole year?
Show working to explain your answer.
Sorry! I just had tea.
- bob bundy
- 2012-12-10 06:07:58
I will help you but pleae don't keep logging out or this could take ages.
What does the question actually ask you to work out?
Bob
- bob bundy
- 2012-12-10 06:00:52
No. that doesn't make sense. An average should be representative of the values in the table and so somewhere between the lowest and highest values.
Do I assume that you have been asked to work out the average for A and the average for B?
Bob
- ihtemthsbtneedtoasksmthng
- 2012-12-10 05:57:55
yes it is! i think ive got the answer now? is it a= 160cm and b= 150cm?