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Mathematics: Post your doubts here!

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U are considering that there are two chemicals only that should be placed on the four positions.but when we are arranging the rest we are placing two more on those 4 places!
please elaborate with calculation/example if you can (that is, where am i going wrong in my approach)
 
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Oct nov 13 varient 61 q#6 part ii) and iii)
Im not that good in permutation and combination so a detailed explanation will be better ....thanks for ur help!!!image.jpg
 
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p3 please solve daredevil asd and others.............................View attachment 40236
angle aoc= A+A+X=pie (theta=A)
aoc-pie-2A

findAC
cos A=adjacent/r
AC=2rcosA

area of sector ABC
(0.5*(2RcosA)^2*A)*2
u should get
4r^2cos^2ArA

now find the area which of the small thingy (idk what u call it)
(area of sector minus area of triangle)*2 ....because there are two of them
so
2[0.5*r^2*pie-2A)-(0.5*r*2rcosA*sinA)]
simplify and u should get
r^2(pie-2A-sin2A)

now add both the areas and equate to 0.5pier^2 and simplify it
hope it helps! :)
 
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M/J_2005_P6 Q6 (ii) Please someone help me
we know
P((1.9-b)<x<(1.9+b))=o.8 and mean=1.9 and SD=o.15
P((1.9-b-1.9)/0.15)<((1.9+b-1.9)/0.15) = o.8
ph of (b)/0.15)+ ph of (+b)/0.1 - 1= o.8
2ph of (b)/0.15=1.8
ph of (b)/0.15=0.9
from the normal distribution graph find the z value for o.9 p if you see the small ttable giveen below it would be easier z= 1.282

(b)/0.15=1.282
b=0.1923
limits 1.9-b and 1.9+b
that is 1.7077 and 2.0923
round of to 3 decimal places
 
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