Author Topic: Standard Math Q+A Thread (Read 254174 times) Tweet Share

RuiAce · « **Reply #360 on:** August 09, 2017, 01:55:53 pm »

Quote from: emilybrooks99 on August 09, 2017, 01:48:53 pm

Hey Could someone please help me with these questions and let me know which formulas to use. Thanks!

$\text{The sine rule and the angle sum of a triangle}\\ \text{will prove useful to find }CD.$
$\text{Using the angle sum of }\triangle BCD\\ \angle CBD = 180^\circ - 87^\circ - 38^\circ = 55^\circ$
$\text{Then, it follows from the sine rule that}\\ \frac{CD}{\sin 50^\circ} = \frac{30}{\sin 38^\circ}\\ \text{Which you can just rearrange and put into your calculator}$

EEEEEEP · « **Reply #361 on:** August 09, 2017, 02:31:17 pm »

Quote from: emilybrooks99 on August 09, 2017, 01:48:53 pm

Hey Could someone please help me with these questions and let me know which formulas to use. Thanks!

First Q
- Inflation is the rise in cost of certain products or services in a year.
1. Annualise the amount
2. Use the inflation rate to calculate the amount in 3 years (annual amount)
3. Divide it by 12
..
edit nvm.

RuiAce · « **Reply #362 on:** August 09, 2017, 04:08:22 pm »

Isn't the first one just multiplication by 1.026³?

bridie_2345 · « **Reply #363 on:** August 12, 2017, 11:10:30 am »

Hi there, I'm unsure of question 5. The answer is D through 1/13 X1/13 but I don't understand why it can't be 13/52 X 13/52

Thanks!

RuiAce · « **Reply #364 on:** August 12, 2017, 11:22:07 am »

Quote from: bridie_2345 on August 12, 2017, 11:10:30 am

Hi there, I'm unsure of question 5. The answer is D through 1/13 X1/13 but I don't understand why it can't be 13/52 X 13/52

Thanks!

13 is the number of cards in a suit in the deck of cards. So if you wanted to draw, say, two spades, then it would be 13/52 * 13/52.

You want a specific rank (aka. value) for the card. There's only 4 of those; one in each suit, so we have 4/52 * 4/52.

Juliesokha · « **Reply #365 on:** August 13, 2017, 12:32:39 pm »

Hello everyone,

I have my trails tomorrow (I know my school is weird but yea) and I need HELP - anyways I have given up on this probability question so bad rn, so can anyone please help me with it?

In a group of 32 people at a party, 20 are male and of those 2 are vegetarians. A third of the females at the party are vegetarians. A person is chosen at random from those at the party. What is the probability that they ARE NOT A VEGETARIAN?
a). 3/16
b). 3/8
c). 13/16
d). 15/16

Thank you <3

RuiAce · « **Reply #366 on:** August 13, 2017, 12:47:48 pm »

Quote from: Juliesokha on August 13, 2017, 12:32:39 pm

Hello everyone,

I have my trails tomorrow (I know my school is weird but yea) and I need HELP - anyways I have given up on this probability question so bad rn, so can anyone please help me with it?

In a group of 32 people at a party, 20 are male and of those 2 are vegetarians. A third of the females at the party are vegetarians. A person is chosen at random from those at the party. What is the probability that they ARE NOT A VEGETARIAN?
a). 3/16
b). 3/8
c). 13/16
d). 15/16

Thank you <3

$\text{Break down step by step}\\ \text{Splitting the genders, we have}\\ \text{20 male, 12 female}$
$\text{We are told 2 males are vegeterians.}\\ \text{Now, }\frac13\text{ of 12 is 4}\\ \text{This tells us that 4 females are vegeterians.}$
$\text{Hence, in total we have }6\text{ people who are vegeterian}$
$\text{So since we have 32 in total, we have 26 people that are }\textbf{not}\text{ vegeterian}\\ \text{So the required probability is }\frac{26}{32} = \frac{13}{16}$

Juliesokha · « **Reply #367 on:** August 15, 2017, 11:47:32 am »

Hello,

Also, this one as well.

Lauren owns 400 $3 ordinary shares and 300 $2 preferences shares. The current prices of the ordinary shares and preferences shares are $4.20 and $3.60 respectively. The dividend on the ordinary shares is 55c and on the preference shares is 3%. Calculate total dividend.

EEEEEEP · « **Reply #368 on:** August 15, 2017, 12:05:37 pm »

Quote from: Juliesokha on August 15, 2017, 11:47:32 am

Hello,

Also, this one as well.

Lauren owns 400 $3 ordinary shares and 300 $2 preferences shares. The current prices of the ordinary shares and preferences shares are $4.20 and $3.60 respectively. The dividend on the ordinary shares is 55c and on the preference shares is 3%. Calculate total dividend.

Let's gather some info.
1. 400 ord shares,
2. 300 Pref shares

Lets calculate dividends.
1. 400 x 0.55 = 220
2. 300 x (0.03 x $3.60) = 32.4

Total = 252.4

Juliesokha · « **Reply #369 on:** August 15, 2017, 08:42:32 pm »

My goodness this looks sooooo easy but didn't we have to use the formula Dividend = Dividend Yield x Market Value?

EEEEEEP · « **Reply #370 on:** August 15, 2017, 08:47:32 pm »

Quote from: Juliesokha on August 15, 2017, 08:42:32 pm

My goodness this looks sooooo easy but didn't we have to use the formula Dividend = Dividend Yield x Market Value?

Yes.

WHich is what I did here (apologies for skipping a step

)
" 300 x (0.03 x $3.60) = 32.4"
I multiplied the bracket by 300 as we had 300 preference shares =)
300 = no. pref shares
0.03 = the yield/percentage
3.6 = market price

Aaron2230 · « **Reply #371 on:** August 25, 2017, 02:02:19 pm »

it is a question from my schools past exam papers, the question is on bearings but there is no solution for me to look at as it is a multi choice question.

RuiAce · « **Reply #372 on:** August 25, 2017, 02:06:50 pm »

Quote from: Aaron2230 on August 25, 2017, 02:02:19 pm

it is a question from my schools past exam papers, the question is on bearings but there is no solution for me to look at as it is a multi choice question.

$\text{It }\textit{may}\text{ help to draw an arrow pointing}\\ \text{from X to Y, and from Z to Y}\\ \text{as these are the relevant bearings.}$
$\text{The line pointing to the right from X implies that this is the bearing }090^\circ\\ \text{It appears that to go from X to Y, we must add }10^\circ\\ \text{So the bearing of Y from X is }100^\circ.\\ \textbf{Note that we added as we had to go clockwise}$
$\text{The line pointing to the left from Z implies that this is the bearing }270^\circ\\ \text{It appears that to go from X to Y, we must subtract }10^\circ\\ \text{So the bearing of Y from X is }260^\circ.\\ \textbf{Note that we subtracted as we had to go anti-clockwise}$
$\therefore \boxed{A}$

Aaron2230 · « **Reply #373 on:** August 25, 2017, 02:11:00 pm »

thanks

Aaron2230 · « **Reply #374 on:** August 25, 2017, 02:48:36 pm »

need help with a question, its somewhat simple just wanted to know ways to do it easy.
The question is the image below.

Search the forums now!

We have moved!

Author Topic: Standard Math Q+A Thread (Read 254174 times) Tweet Share

RuiAce

Re: General Math Q+A Thread

EEEEEEP

Re: General Math Q+A Thread

RuiAce

Re: General Math Q+A Thread

bridie_2345

Re: General Math Q+A Thread

RuiAce

Re: General Math Q+A Thread

Juliesokha

Re: General Math Q+A Thread

RuiAce

Re: General Math Q+A Thread

Juliesokha

Re: General Math Q+A Thread

EEEEEEP

Re: General Math Q+A Thread

Juliesokha

Re: General Math Q+A Thread

EEEEEEP

Re: General Math Q+A Thread

Aaron2230

Can someone help with a question please

RuiAce

Re: Can someone help with a question please

Aaron2230

Re: Can someone help with a question please

Aaron2230

Help with a question again plz

Recent Posts

User:
Password: