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secondary 4 | E Maths
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Yos
Yos

secondary 4 chevron_right E Maths chevron_right Singapore

The answer for the two values are 10 and 11. However, are any two numbers (besides the two numbers mentioned earlier) acceptable as long as it adds up to 21 with the mean mark of 10 and mode = 11 such as 19 and 2 or 15 and 6?

Date Posted: 3 years ago
Views: 290
AC Lim
AC Lim
3 years ago
Yes.
15 and 6 or
19 and 2 or
18 and 3 or
17 and 4

Except
16,5
14,7
13,8
12,9
Yos
Yos
3 years ago
Why 16 and 5 is not acceptable although it meets all the criterias?
AC Lim
AC Lim
3 years ago
Oh sorry, 16 and 5 is accepted. I am wondering 21 and 0 acceptable?
Yos
Yos
3 years ago
I have no clue either. Well I just have to discuss this answer with my math teacher tomorrow. Anyways thanks for your confirmation
Eric Nicholas K
Eric Nicholas K
3 years ago
I did not think of 21 and 0...

That was something I missed out
Eric Nicholas K
Eric Nicholas K
3 years ago
Unless the “unclear part” really meant “physically unclear to the eye”.

Because if this is the case, the most likely case of “unclear writing” are two-digit numbers (like those type of questions where ink is spilled on the paper thus covering the ones digit of each mark).

Other than that, I don’t see why other solutions cannot be accepted. All are possible.
AC Lim
AC Lim
3 years ago
@eric your are right. Question should be find the 2 digit missing numbers. Else, all are acceptable.

See 3 Answers

Based on my understanding,
Mode : value with highest frequency, so i assume you need to have value 11 more.

So with that since u find one value , you can find the other value
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Nathaniel
Nathaniel's answer
30 answers (Tutor Details)
1st
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Nathaniel
Nathaniel's answer
30 answers (Tutor Details)
Yaaa thats about it , so u cannot really have other solutions
Eric Nicholas K
Eric Nicholas K
3 years ago
No, many solutions are possible. Depending on the values chosen, many of them fits the “11 as the mode” condition.

It does not have to be three 11s to be counted as a mode in this case. So, I suspect the question is largely flawed (but I know the intention of 11 and 10 being the solution)
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Definitely a problem with the question (multiple solutions possible).