Nabteb 2016 verified Further Mathematics questions and answers (expo)

12a)
tabulate
Marks| 1-10, 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, 81-90, 91-100
F| 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
C.B| 0.5-105, 10.5-205, 20.5-305, 30.5-405,
40.5-505, 50.5-605, 60.5-705, 70.5-805,
80.5-905, 90.5-1005
C.F| 0+3=3, 3+17=20, 20+41=61, 61+85=146, 146+77=243, 243+115=358, 358+101=459, 459+64=523, 523+21=544, 544+6=550

No 12 graph diagram



6a)
(1+4+k+k+4+11)/5=k+1
(20+2k)/5=k+1
20+2k=5(k+1)
20+2k=5k+5
20-5=5k-2k
15=3k
k=15/3
k=5
6b)
1,4,5,9and 11
TABULATE
x:1,4,5,9,11
x-x^-:-4,-1,0,4,6
(x-x^-)^2:16,1,0,16,36
E(x-x^-)^2=69
Hence=E(x-x^-)^2/n
=69/5
=13.8
2)
(5,2)(-4,k)(2,1)
(y3-y2)/(x3-x2)=(y2-y1)/(x2-x1)
(1-k)/2-(-4)=(k-2)/(-4-5)
(1-k)/(2+4)=(k-2)/-9
(1-k)/6=(k-2)/-9
-9(1-k)=6(k-2)
-9+9k=6k-12
9k-6k=-12+9
3k=-3
k=-1
7)
m1=3
u1=8m/s
m2=?
u2=5m/s
v=6m/s
m1u1+m1u2=(m1+m2)v
3*8+m2*5=(3+m2)6
24+5m2=18+6m2
24-18=6m2-5m2
m2=6
5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60
5b)
pr(only age and fully gained admission)=4/5*3/4*1/3
=1/5
4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11
Therefore m=5, n=7
3a)
If f(x+2)=6x^2+5x-8)
To find f(5)
Therefore f(x+2)=f(5)
where x+2=5
x=5-2
x=3
therefore f(5)=6(3)^2+5(3)-8
=6(9)+15-8
=54+7
=61
3b)
(7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3)
(24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3)
(48+26root6+18)/(32-12)
=(66+26root6)/20
=66/20+(26root6/20)
=33/10+(13root6/10)
=3.3+1.3root6

13a)
M=2=mathematics=2
P=5=physics=5
C=3=chemistry=3
i)
M.P.C=2! x 5! x3! =2*1*5*4*3*2*1*3*2*1
=2*120*6=1,440ways
ii)
p=5!
=5*4*3*2*1
=120ways

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Post Summary

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Article Name	Nabteb 2016 verified Further Mathematics questions and answers (expo)
Description
Author Name	Head Admin
Published On	5/13/2016
Post Category	education

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