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Box and Wisker (1 Viewer)

Paj20

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Does the interquartie range in box and wiskers represent 50% of the data.. and each side of the median in that range 25%?
 

Paj20

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-tal- said:
solution and working + diagram in the attachment.
thanks heaps... you dont happen to have the answers to the whole thing do u??


and in that.. does that mean angle A is 18 degrees?
 

-tal-

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Paj20 said:
thanks heaps... you dont happen to have the answers to the whole thing do u??


and in that.. does that mean angle A is 18 degrees?
No, it doesn't, because you have to find the small angle that A makes with C, then add that to the 18.

Do you want me to find A?

Hmm.. maybe it's better I look at the question again.
 
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Paj20

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-tal- said:
No, it doesn't, because you have to find the small angle that A makes with C, then add that to the 18.

Do you want me to find A?

hmmm i thought it was just 90-72 because it is on the north
 

-tal-

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ii) You can use the cosine rule here.

AC² = AB² + BC² - 2(AB)(BC)CosB
AC² = 300² + 400² - 2(300)(400)Cos 96
AC² = 275086.8312
AC = 524m (nearest m)

iii) Then, since you have AC, you can use the sine rule.

sin 96/524 = sin BCA/400
400 sin 96/524 = sin BCA
BCA = 49 degrees, 24 min (nearest min)

Forgot to find bearing.

Sorry for the long reply. Went to eat.
 
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-tal-

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Paj20 said:
How did you get blue to equal 49 24' !!?????

iii) Then, since you have AC, you can use the sine rule.

sin 96/524 = sin BCA/400
400 sin 96/524 = sin BCA
BCA = 49 degrees, 24 min (nearest min)
 

Paj20

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-tal- said:
iii) Then, since you have AC, you can use the sine rule.

sin 96/524 = sin BCA/400
400 sin 96/524 = sin BCA
BCA = 49 degrees, 24 min (nearest min)

ohh yes... i see.. i used the cosine rule because there was 3 sides


kk Thanks.:rolleyes:
 

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