Standard Math Q+A Thread

[RuiAce]

Hi all

So I am currently prepping for my general maths exam and am stuck on this question. Any help would be appreciated.

Josephine invested $1000 at the end of each year for five years. Her investment earned interest at 4.8% per annum compounded annually. What was the total of Jospehine's investment (to the nearest dollar) at the end of the fifth year?

Thanks 

Did this question come from a paper prior to the syllabus change in 2014? Back then they had a formula to handle the value of an annuity, but it has since been scrapped.

[morning_sunshine]

Did this question come from a paper prior to the syllabus change in 2014? Back then they had a formula to handle the value of an annuity, but it has since been scrapped.

Yes it has come from a past paper from 2001, but was unsure of whether this was something I needed to know. So is this something i should not worry about or is there now a new way to work it out?

[RuiAce]

_____________________

[jamonwindeyer]

Yes it has come from a past paper from 2001, but was unsure of whether this was something I needed to know. So is this something i should not worry about or is there now a new way to work it out?

Just to clarify, although you could use Rui's method above, you'd never be asked to do it in an exam like that - No stress!! Always tough to know with old papers so glad you asked

[morning_sunshine]

thank you guys so so much I was really struggling so your help is so appreciated! I will be posting way more questions later   
but thanks again for the amazing help !! 

[gaudy]

QUESTION:
The hsc marks for a particular course are normally distributed with a mean of 65 and a standard deviation of 8.
What percentage of results lie between 65 and 73?

pls help

[jakesilove]

QUESTION:
The hsc marks for a particular course are normally distributed with a mean of 65 and a standard deviation of 8.
What percentage of results lie between 65 and 73?

pls help

Hey! On your formula sheet, it says 'Approximately 68% of scores have z-scores between -1 and 1". This means that, one standard deviation away from the medium to either side contains 68% of the results! However the question only is interested in those scores ABOVE the medium, not to either side. So, we divide the percentage of scores by two, getting 34% as our answer

Jake

[Potatohater]

Hey, how would I go about solving this question?
Mary is designing a website that requires unique login generation. She plans to generate logins using 2 capital letters followed by a series of numerals 0-9 inclusive, all logins have the same number of numerals, repitition is allowed, what is the minimum number of numerals needed for each login so that Mary can generate at least 3 million logins?

[stephanieazzopardi]

Hey, how would I go about solving this question?
Mary is designing a website that requires unique login generation. She plans to generate logins using 2 capital letters followed by a series of numerals 0-9 inclusive, all logins have the same number of numerals, repitition is allowed, what is the minimum number of numerals needed for each login so that Mary can generate at least 3 million logins?

Hi Potatohater,

I think this is a case of guess and check. So the first two items of her password are 2 letters. So the number of possible combinations would equal 26 x 26. To work out how many numerals she would need to have in her password to generate at least 3 million logins, you would just type 26 x 26 x 10 x 10 x 10 etc, and see how many sets of x 10 you had in your calculation before you reached at least 3 million, each x 10 representing a numeral.

So if we do 26 x 26 x 10 x 10 x 10 (3 numerals), we get 676 000 = not at least 3 million, right?
So let's do 26 x 26 x 10 x 10 x 10 x 10 (4 numerals). We get 6 760 000 = at least 3 million.

So the answer should be 4. Hope that helps! Let us know if you have any more questions

[RuiAce]

Remark: It's doable with logs but I doubt that's in general so I'd vouch for guess and check as well

[stephanieazzopardi]

Remark: It's doable with logs but I doubt that's in general so I'd vouch for guess and check as well

No logs in General

[morning_sunshine]

hey guys
So i am doing a past trial paper and got stuck on this question.

It has worked solutions as seen in the picture but I still don't understand how each step was reached.. if that makes sense?
I just need some help understanding the question or if there is a simpler way to do this question let me know please!

[Shadowxo]

hey guys
So i am doing a past trial paper and got stuck on this question.

It has worked solutions as seen in the picture but I still don't understand how each step was reached.. if that makes sense?
I just need some help understanding the question or if there is a simpler way to do this question let me know please!

This is using the cosine rule
a²=b²+c²+2bc cosA
a = BC, b = AC c= AB
Or in other words a= 90, b = c = x
A= 113º
That's how they get the first line
From there they just rearrange it. They get all the terms with x on one side (they already are though), all the terms without on another, then they factor out the x², get x² as the subject then square root both sides

If you're unsure about the cosine rule, you should brush up on it (don't know what's required for your course though) and if you still need help let me know. Rest is just rearranging the equation to make x the subject

[Potatohater]

Hey, so I tried this question 2 different ways so far and none of them have worked, so how should I go about this question:
2 cities lie on the same line of longitude and are 4356km apart. One city is located 47 degrees north, what is the latitude of the 2nd city if it is south of the 1st?

[RuiAce]

Hey, so I tried this question 2 different ways so far and none of them have worked, so how should I go about this question:
2 cities lie on the same line of longitude and are 4356km apart. One city is located 47 degrees north, what is the latitude of the 2nd city if it is south of the 1st?

I'm not entirely confident of this method (mostly because I didn't actually do general maths), but I believe you should use the arc length formula \(l=\frac{\theta}{360}2\pi r\) (from the formula sheet) to figure out the angle between the two cities. Note that \(l=4356\), and of course \(r=6400 \)

Then, you know that the first city is at 47 degrees N. So subtract whatever \(\theta\) was from 47 N to find the latitude of the second city.