Author Topic: Standard Math Q+A Thread (Read 182549 times) Tweet Share

RuiAce · « **Reply #585 on:** April 01, 2018, 02:21:26 pm »

Quote from: Razeen25 on April 01, 2018, 01:45:41 pm

I’m actually confused. How do you calculate the cumulative frequency from a stem and leaf plot?

The numbers less than or equal to 163 in that stem and leaf plot are
151, 151, 151, 152, 160, 160, 163, 163, 163, 163.

There are 10 numbers. So the cumulative frequency of 163 is 10.

Quote from: Lumenoria on April 01, 2018, 02:07:10 pm

Okay, so this is the solution on the answers for a past trial paper I've been doing, but I swear to god it's wrong - or I'm just really stupid. Can someone please confirm? My answer is waaay off this, and I swear I haven't done anything wrong because I've gotten the same answer three times. I don't understand why they've minused the a and the b when the formula clearly states you're supposed to add them?

Presumably, the area enclosed by the trapezium but excluding that triangle was meant to be shaded. Judging by the quality of the picture, it looks like that the shading disappeared when the image got scanned.

The question most likely only wanted the area of a "shaded" region. So if the shaded region was the trapezium without the triangle in the middle, then their working out is fine

Lumenoria · « **Reply #586 on:** April 02, 2018, 01:54:31 pm »

Can someone please help me with the last 3 of question 8? I am literally lost. The answers are;
c. 9/100
d. 0.066
e. 0.624
I need working out aswell, because I literally have no clue how they've ascertained such answers and my exams are tomorrow.

Thankyou so much!!

RuiAce · « **Reply #587 on:** April 02, 2018, 02:18:09 pm »

Quote from: Lumenoria on April 02, 2018, 01:54:31 pm

Can someone please help me with the last 3 of question 8? I am literally lost. The answers are;
c. 9/100
d. 0.066
e. 0.624
I need working out aswell, because I literally have no clue how they've ascertained such answers and my exams are tomorrow.

Thankyou so much!!

$\text{We know from part a) that the probability it rains on}\\ \text{any particular day is }\frac{9}{30}\text{, which simplifies down to }\frac{3}{10}.$
$\text{In part c), all we're saying now is that we want a specific }\textbf{two days}\text{ to be raining}.\\ \text{So this is just }\frac{3}{10}\times \frac{3}{10},\text{ or alternatively }\left( \frac{3}{10} \right)^2.\\ \text{Which is }\frac{9}{100}$
$\text{Similarly, because the probability it doesn't rain on a particular}\\ \text{day in October is }\frac{18}{31}, \\ \text{if we want no rain on }\textbf{5}\text{ days then the answer is}\\ \left( \frac{18}{31} \right)^5$
_________________________________________________
$\text{Whereas for e), we should think about the complement again.}\\ \text{What we }\textbf{don't}\text{ want is for both days of the}\\ \text{weekend to not be rainy.}$
$\text{Since the probability of it not raining in December is }\frac{19}{31},\\ \text{what we require is }1 - \left( \frac{19}{31} \right)^2$

Lumenoria · « **Reply #588 on:** April 02, 2018, 06:37:37 pm »

Quote from: RuiAce on April 02, 2018, 02:18:09 pm

$\text{We know from part a) that the probability it rains on}\\ \text{any particular day is }\frac{9}{30}\text{, which simplifies down to }\frac{3}{10}.$
$\text{In part c), all we're saying now is that we want a specific }\textbf{two days}\text{ to be raining}.\\ \text{So this is just }\frac{3}{10}\times \frac{3}{10},\text{ or alternatively }\left( \frac{3}{10} \right)^2.\\ \text{Which is }\frac{9}{100}$
$\text{Similarly, because the probability it doesn't rain on a particular}\\ \text{day in October is }\frac{18}{31}, \\ \text{if we want no rain on }\textbf{5}\text{ days then the answer is}\\ \left( \frac{18}{31} \right)^5$
_________________________________________________
$\text{Whereas for e), we should think about the complement again.}\\ \text{What we }\textbf{don't}\text{ want is for both days of the}\\ \text{weekend to not be rainy.}$
$\text{Since the probability of it not raining in December is }\frac{19}{31},\\ \text{what we require is }1 - \left( \frac{19}{31} \right)^2$

Omg thankyou so much, seriously!! This clarified things so much, thankyou!!

joemassoud · « **Reply #589 on:** April 04, 2018, 12:35:05 pm »

Can someone please help me with this question

RuiAce · « **Reply #590 on:** April 04, 2018, 12:36:50 pm »

Quote from: joemassoud on April 04, 2018, 12:35:05 pm

Can someone please help me with this question

\begin{align*}A&=\pi r (r+s)\\ \frac{A}{\pi r}&= r + s\\ \frac{A}{\pi r} - r &= s\end{align*}
\[\therefore \boxed{s = \frac{A}{\pi r} - r} \]

joemassoud · « **Reply #591 on:** April 04, 2018, 12:50:14 pm »

Quote from: RuiAce on April 04, 2018, 12:36:50 pm

\begin{align*}A&=\pi r (r+s)\\ \frac{A}{\pi r}&= r + s\\ \frac{A}{\pi r} - r &= s\end{align*}
\[\therefore \boxed{s = \frac{A}{\pi r} - r} \]

Thank you soo much!

Just one more question, is there a 'rule of thumb' in regards to what you must do first when changing the formula of a subject.? (in a particular order)

MAGGOT · « **Reply #592 on:** April 04, 2018, 03:44:26 pm »

Quote from: joemassoud on April 04, 2018, 12:50:14 pm

Thank you soo much!

Just one more question, is there a 'rule of thumb' in regards to what you must do first when changing the formula of a subject.? (in a particular order)

in reverse of bodmas form

Razeen25 · « **Reply #593 on:** May 07, 2018, 08:10:53 pm »

Hi,
Could someone please help me with this question? I asked my teacher but I still don’t understand it : (.

“James walked 2.8 km due west of home, then turned and walked on a bearing of 296° until he was 4.9 km from home. What is James’s bearing from home?”

The answer is 281°, but I keep getting 303° SMHH.

RuiAce · « **Reply #594 on:** May 07, 2018, 08:38:32 pm »

Quote from: Razeen25 on May 07, 2018, 08:10:53 pm

Hi,
Could someone please help me with this question? I asked my teacher but I still don’t understand it : (.

“James walked 2.8 km due west of home, then turned and walked on a bearing of 296° until he was 4.9 km from home. What is James’s bearing from home?”

The answer is 281°, but I keep getting 303° SMHH.

(May need to click on the image.)

Razeen25 · « **Reply #595 on:** May 07, 2018, 08:56:48 pm »

Quote from: RuiAce on May 07, 2018, 08:38:32 pm

(May need to click on the image.)
(Image removed from quote.)

YOU EXPLAINED THIS PERFECTLY. Thank you soo much!

Lumenoria · « **Reply #596 on:** May 09, 2018, 08:05:22 am »

15 and 16 please!

RuiAce · « **Reply #597 on:** May 09, 2018, 09:11:08 pm »

Quote from: Lumenoria on May 09, 2018, 08:05:22 am

15 and 16 please!

$\text{The back seat and front seat makes no difference}\\ \text{for the sake of 15a.}\\ \text{The question is essentially done the same way as arranging 4 people in a line.}$
The reason why it makes no difference is because we haven't specified any conditions on who sits at the front or who sits at the back.
$\text{So it'll just be }4\times 3\times 2\times 1\text{, or if you prefer, }4!.$
$\text{For 15b, if Kramer drives then he cannot sit at the back.}\\ \text{So in the back, we have either J and G, G and E or J and E.}$
_________________________________________
$\text{Again, for 16a, this is just arranging 5 people in a line}\\ \text{so it'll be }5\times 4\times 3\times 2\times 1\text{, or if you prefer, }5!.$
$\text{Whereas a restriction comes into play for the sake of 16b.}\\ \text{The fact that the three girls must be in front of the two boys deadlocks their positions a bit.}$
$\underbrace{\boxed{\times \quad \times \quad \times}}_{\text{girls}}\quad \underbrace{\boxed{\circ \quad \circ}}_{\text{boys}}\\ \text{We force the 3 girls to occupy the first three spots, and the boys the last 2,}\\ \text{and then we arrange them }\textbf{only among themselves}.\\ \text{i.e. we no longer care about arranging girls and boys between each other.}$
$\text{Which, is just arranging 3 in a line, and then 2 in a line immediately after.}\\ \text{So }3!\times 2!.$

emilyygeorgexx · « **Reply #598 on:** May 09, 2018, 10:42:54 pm »

Quote from: Lumenoria on May 09, 2018, 08:05:22 am

15 and 16 please!

If you are doing ordered combinations, you can make your life easier by using the permutations function on the calculator (shift, multiplication sign). If you are doing unordered combinations you can use the combinations function (shift, divide sign).

RuiAce · « **Reply #599 on:** May 09, 2018, 11:05:25 pm »

Quote from: emilyygeorgexx on May 09, 2018, 10:42:54 pm

If you are doing ordered combinations, you can make your life easier by using the permutations function on the calculator (shift, multiplication sign). If you are doing unordered combinations you can use the combinations function (shift, divide sign).

Whilst this is true, I don't think this is relevant for the questions provided. They don't involve selecting a certain amount of people out of a group for something here, they're just arranging every person present into positions.

Of course, for the section coming right up, combinations can be useful.

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