HELP Any mathematicans out there (1 Viewer)

[#23]

1. What is an outlier. How will I use it if given a question on it?
2. Simpsons Rule. This rectangular region contains a dam and a field. All measurements are metres.
a) use simpsons rule twice to find the approx. area of the field (easy)
b) hence calcualte the area covered by the dam (hard)
Anyone knw how 2 do the b). Below is a quick sketch of what the pic looks like lol

Ok so its a BIG rectangle, abit more than half of it is filled with DAM.


<--------------- 120 ---------------->
DAMMMMMMMMMMMMMMMMMM Going downwards from top to bottom its 110
DAMMMMMMMMMMMMMMMMMM
all damn

DAMMMMMMMMMMMMMMMMMM
DAMMMMMMMMMMMM
DAMMMMMMM

From left to right; First line on left going up covers 80 up to the 3rd 'DAM' word, then 62 (new line goingup) to the 2nd 'dam word, 55 going to the middle, then 78 to 2nd dam and to the 3rd dam word its 85. NOW the h/3, theres 4 of those boxes things so 120/4 = 30 each.

PS. Answers to a and b are
a) 8350
b) 4850


Thanks alot

 

[100percent]

1. What is an outlier. How will I use it if given a question on it?

outlier is the weird one, eg, what people got on their math test on monday, person A got 50%, person B got 45%, person C got 48%, person D got 51% and person E got 99%, person E is therefore the outlier.
i don't do general maths but i'm guessing you ignore any outliers?

i don't understand your simpson rule question clearly, prehaps you can scan a pic and post it up?

 

[#23]

Here is the link for the 2 questions Im having problems with. 6 (b) and 7 (c) http://img452.imageshack.us/img452/5926/sdas2jv.jpg

Also, for the topic staging multiple events, the formulas 1 - the opposite technique.. What exactly are you supposed to minus.
Five men and 2 women have volunetered tow ork on a 3-person commitee which is going to be selected at radnom. what is the probability the commitee will consist of:
at most 2 men?
How can you use opposite tech for that

Thanks alot

 

[100percent]

ah i see, your simpsons rule question, you said part a is easy so i won't bother, basically, the area you found is the non water part of the rectangle, so the water part is area of rectangle minus area you found in 'a'

area of rectangle 120x110 = 13200
area you found in part 'a' = 8350

area of rectangle minus area found in part 'a' = 4850

 

[100percent]

Also, for the topic staging multiple events, the formulas 1 - the opposite technique.. What exactly are you supposed to minus.

umm, how can i explain it. hmm. lets use an example. some person shoots an arrow and as a probability of 1/3 that he/she will hit the target.
find the probability that in 10 shoots, he/she hits the target ATLEAST once. so normally, you would find probability of hitting exactly once, then exactly twice then exactly 3 times and so on. this is a stupid way to do it, an easier way is to find the probablity of he/she not hitting the target at all, so you find the probability that he/she hits the target zero times. so you do 1 minus zero times probability, and thats your answer.

Five men and 2 women have volunetered tow ork on a 3-person commitee which is going to be selected at radnom. what is the probability the commitee will consist of:
at most 2 men?
How can you use opposite tech for that

at most 2men, so it can't have 3men, so find the prob of it having 3men and then do one minus the probability you just found.
so probability of having 3men
5C3/7C3 => 10/35 => 2/7

now you do 1 - 2/7 so that is equal to 5/7

 

[#23]

100percent, cheers, thanks alot man.. answers were great, now i understand

 

[#23]

Woops, I got afew more questions.

http://img470.imageshack.us/img470/7885/q101kb.jpg
Question 10 b, c and d

&

http://img456.imageshack.us/img456/3228/q17ro.jpg
Question 1 a b and c

Thanks alot ))))

 

[100percent]

10b)
let '<' be the angle sign
<OTB= 90°-23° = 67°
cos (<OTB) = OT/TB
cos 67°= 85/x
x=85/cos 67°
x=217.5m

10c)
<OTB=67°
<TOB=90°
<OBT=90-67=23°
then <TBA=90-23=67°
(or if you can just say <OTB=<TBA since alternate <, TO||AB)
<BTA=23-20=3°
therefore <BAT=180-(3+67)=110 (angle sum of triangle)

10d)
sin 110/217.5=sin 3/x
x*sin 110=217.5sin 3
x=[217.5sin3]/[sin 110]
x=12.11m

1a)
alpha = 180-31=149°
beta = 180 - (149+18)=13°

1b)
72/sin B = x/sin A
72sin 149 = xsin 13
[72sin 149]/sin 13 = x
x=164.848

1c)
sin 31 = AD/AC
sin 31 = h/x
but x = 164.848 from b
therefore sin 31 = h/164.848
164.848sin 31 = h
h=84.9

alright, i'm going to do 1 last pass paper
Good Luck Tomorrow

 

[#23]

10b)
let '<' be the angle sign
<OTB= 90°-23° = 67°
cos (<OTB) = OT/TB
cos 67°= 85/x
x=85/cos 67°
x=217.5m

10c)
<OTB=67°
<TOB=90°
<OBT=90-67=23°
then <TBA=90-23=67°
(or if you can just say <OTB=<TBA since alternate <, TO||AB)
<BTA=23-20=3°
therefore <BAT=180-(3+67)=110 (angle sum of triangle)

10d)
sin 110/217.5=sin 3/x
x*sin 110=217.5sin 3
x=[217.5sin3]/[sin 110]
x=12.11m

1a)
alpha = 180-31=149°
beta = 180 - (149+18)=13°

1b)
72/sin B = x/sin A
72sin 149 = xsin 13
[72sin 149]/sin 13 = x
x=164.848

1c)
sin 31 = AD/AC
sin 31 = h/x
but x = 164.848 from b
therefore sin 31 = h/164.848
164.848sin 31 = h
h=84.9

alright, i'm going to do 1 last pass paper
Good Luck Tomorrow

Thanks man.. ) (PS. Btw the last 2 were wrong but doesnt matter I managed to work the B out but can't find C hehe Thanks)

 

[100percent]

Thanks man.. ) (PS. Btw the last 2 were wrong but doesnt matter I managed to work the B out but can't find C hehe Thanks)

sorry, i took wrong angle (careless me)
10b)
sin 18/x = sin 13/72
x=72sin18/sin13
x=98.907

10c)(c was wrong coz my x was wrong from previous answer)
sin 31 = h/x
sin 31 = h/98.907
98.907sin 31= h
h = 50.9

 

[#23]

sorry, i took wrong angle (careless me)
10b)
sin 18/x = sin 13/72
x=72sin18/sin13
x=98.907

10c)(c was wrong coz my x was wrong from previous answer)
sin 31 = h/x
sin 31 = h/98.907
98.907sin 31= h
h = 50.9

Thanks alot, thats all right, 1 hour and 10 minutes before the test starts damn

 

[chilena4life]

Need A Maths Tutor For General Maths Urgently For Sydney City Area...get At Me Asp If U Know Any Good Ones Chau Xoxo