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secondary 4 | E Maths
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Hi I cannot comprehend this question at all and the answer from the answer sheet in green someone please explain to me in clear detail on how to do this question and the answer is thank you so much I don’t understand the answer at all like what does it mean 1 of the 3 marks is 29 what does it mean 1 of the 3???
Date Posted: 4 years ago
Views: 206
There are 6 marks in total. Modal mark means mark that appears the most times.
Since it is 29, and there is already one 29, 25 and 26, the mark 29 must appear more than once in order to be the modal mark.
Therefore, for the remaining 3 marks, p,q and r, at least one of them must be a 29.
Since it is 29, and there is already one 29, 25 and 26, the mark 29 must appear more than once in order to be the modal mark.
Therefore, for the remaining 3 marks, p,q and r, at least one of them must be a 29.
I’m sorryyy but I thought it’s the number that appears the most?
Yes edited already.
I am editing my explanation as I type
I am editing my explanation as I type
Since there are 6 numbers (even number), and the median is 28,
The middle 2 numbers (3rd and 4th) must have an average of 28.
Total of the middle two numbers
= 28 x 2 = 56
(Remember that the median for a list with even number of numbers is obtained by finding the average of the middle 2 numbers)
The middle 2 numbers (3rd and 4th) must have an average of 28.
Total of the middle two numbers
= 28 x 2 = 56
(Remember that the median for a list with even number of numbers is obtained by finding the average of the middle 2 numbers)
Since the average is 28, and we know there is at least one 25, one 26 and at least two 29s,
We find the total of all 6 scores first and subtract 25,26, 29 and 29 to find the sum of the two remaining two numbers.
28 x 6 = 168
168 - 25 - 26 - 29 - 29 = 59
The remaining numbers have a total of 59.
We find the total of all 6 scores first and subtract 25,26, 29 and 29 to find the sum of the two remaining two numbers.
28 x 6 = 168
168 - 25 - 26 - 29 - 29 = 59
The remaining numbers have a total of 59.
We have 25,26,29,29 and two unknown numbers, ? and ?
Now, how to find two numbers that add up to 59, and satisfy the condition of the two middle numbers having a total of 56?
Now, recall that the question says p < q < r , which means the numbers are all different.
Since one of them is 29 as we have already deduced,
29 can be the biggest (r) , middle number(q) , or the smallest of the three (p)
Now if we let r = 29, then p and q must be smaller than 29 to satisfy the inequality. So they can only be 28 and below.
But if that's the case , their total cannot add up to 59 since the highest you can get is 28 + 27 = 55. So r cannot be 29.
Now if we let p = 29 instead, then q and r must be bigger than 29 to satisfy the inequality.
But if that's the case , their total cannot add up to 59 since the lowest you can get is 30 + 31 = 61. So p cannot be 29.
This leaves us with q = 29
Now, how to find two numbers that add up to 59, and satisfy the condition of the two middle numbers having a total of 56?
Now, recall that the question says p < q < r , which means the numbers are all different.
Since one of them is 29 as we have already deduced,
29 can be the biggest (r) , middle number(q) , or the smallest of the three (p)
Now if we let r = 29, then p and q must be smaller than 29 to satisfy the inequality. So they can only be 28 and below.
But if that's the case , their total cannot add up to 59 since the highest you can get is 28 + 27 = 55. So r cannot be 29.
Now if we let p = 29 instead, then q and r must be bigger than 29 to satisfy the inequality.
But if that's the case , their total cannot add up to 59 since the lowest you can get is 30 + 31 = 61. So p cannot be 29.
This leaves us with q = 29
So now for p and r, their total has to be 59, p must be smaller than 29 and r must be bigger than 29. And this must result in having the two middle numbers adding to 56.
Since r is bigger than 29, it means r is the largest of the 6 numbers as 29 is bigger than p and also bigger than 25 and 26.
25<26<29
p< 29
So in ascending order, 5 of the 6 numbers are 25,26,29,29,r . We don't know where p is yet, though we know it is below 29.
But we now know that 29 definitely is one of the middle numbers since one of the 29 is the 4th number in the list (p can't be the 4th number since it is smaller than 29)
Since the two middle numbers add up to 56, the other number is 56 - 29 = 27
So we have 25,26,27,29,29,r
This tells us p is actually one of the middle numbers (the 3rd of the 6 in ascending order), and p = 27
So now we can find r.
r = 59 - 27 = 32
Therefore p = 27, q = 29, r = 32
Since r is bigger than 29, it means r is the largest of the 6 numbers as 29 is bigger than p and also bigger than 25 and 26.
25<26<29
p< 29
So in ascending order, 5 of the 6 numbers are 25,26,29,29,r . We don't know where p is yet, though we know it is below 29.
But we now know that 29 definitely is one of the middle numbers since one of the 29 is the 4th number in the list (p can't be the 4th number since it is smaller than 29)
Since the two middle numbers add up to 56, the other number is 56 - 29 = 27
So we have 25,26,27,29,29,r
This tells us p is actually one of the middle numbers (the 3rd of the 6 in ascending order), and p = 27
So now we can find r.
r = 59 - 27 = 32
Therefore p = 27, q = 29, r = 32
If you need more clarification let me know.
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Hope this helps
Date Posted:
4 years ago
Modal mean a set of data that appears most often. Example
This set of number:
12 16 17 17 17 18 18 so modal is 17
Median is If the number of observations is odd, the number in the middle of the list is the median.
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest.
Example set of number (odd)
15 16 18 19 22 25 27 so the median is 19
「Left 3 number」19 「right 3 number」 center number is median. ...19
If set of number is even
Example
12 14 13 15 18 19
The center number is between 13 and 15, so median is (13+15)÷2=14
Mean is is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
This set of number:
12 16 17 17 17 18 18 so modal is 17
Median is If the number of observations is odd, the number in the middle of the list is the median.
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest.
Example set of number (odd)
15 16 18 19 22 25 27 so the median is 19
「Left 3 number」19 「right 3 number」 center number is median. ...19
If set of number is even
Example
12 14 13 15 18 19
The center number is between 13 and 15, so median is (13+15)÷2=14
Mean is is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
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