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15b)
Pr(at most 2)=Pr(0)+Pr(1)+Pr(2)
n=1000,Pr=0.05 Pr(2)=0.95
Pr(x)=e^-landa(landa^x)
Pr(0.05)<0.1 and landa=1.000*0.05=50 e^-50(50^0)/o!+e^-50(50^1)/1!+e^-50(50^2)/2! 1/e^50+50/e^50+250.2e^50 =1/e^50[1+50+25/2] ----------------- 12a) w=weight of the plank moment=force*perpendicular distance clockwise moment=Anticlockwise Make A the reference point Pw=2.5m, AM=(2.5-0.8)m =1.7m g=10m/s AB=3.2m,PA=0.8m Convert 30kg to weight =mg=30*10=300N therefore 300*0.8=w*1.7 w=240/1.7=141.18N Distance of its C.G from P=2.5m 12b) For R1=A, R2=B R1=R2=2R1 Let Q be the reference point 300x*141.18*2.5=2R1(4.8+1) 300x+352.95=11.6R1 300x-11.6R1=-352.95---(1) Let P in the reference point 2R1(0.8+4)=141.15*2.5 9.6R1=352.95 R1=352.95/9.6 =36.77N 1a) A=(2 -1)(1 3) |A|=(2*3--1) |A|=6+1 |A|=7 since the determinant of A is not 0 the A is not non singular matrix To find A^-1 A^-1=(3 1)(-1 2) then 1/7(3 1)(-1 2) A^-1=(3/7 1/7)(-1/7 2/7) 2a) 3root2-3root5/5root2+2roo5 conjugate=5root2-2root5 =(3root2-3root5)(5root2-2root5)/(5root2+2root5)(5root2-2root5) =(15*2-6root10-15root10-6*5)/(25*2-10root10+10root10-4*5) =(30-31root10-30)/(50-20) =-31root10/30 2b) cosx=adj/hyp=12/13 then tanx=opp/adj=5/12 siny=3/5 tany=3/4 tan(x+y)=tanx+tany/1-tanxtany =(5/12+3/4)/(1-(5/12)(3/4) =(14/12)/(1-15/48) =36/125+18/125+16/125 =70/125 =14/25 3a) Pr(A passed)=4/5,pr(A failed)=1/5 Pr(B passed)=3/5,Pr(B failed)=2/5 Pr(C passed)-2/5,Pr(C failed)=3/5 Pr(atleast one passed) =4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5 =24/125+9/125+4/125 =47/125 3b) Pr(atleast one failed) =4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5 =101/125 4) (3+2y)^5 with (a+b)^5 shows a=3,b=2y Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5 and 1 (3+2y)^5 =(3)^5+5(3)^4(2y)+10(3)^3(2y)^2+10(3)^2(2y)^3+5(3)(2y)^4+(2y)^5 =243+810y+1080y^2+720y^3+240y^4+32y^5 (3.04)^5=(3+0.04)^5 =3^5+5(3)^4(0.04)+10(3)^2(0.04)^2+10(3)^2(0.04)^3+5(3)(0.04)^4+(0.04)^5 =243+5*81(0.04)+10*27(0.0016)+10*9(0.000064)+15(0.0000026)+0.0000001 =243+16.2+0.432+0.00576+0.000039+0.0000001 =259.63783 11a) 5x^2-2x+3=0 Divide both sides by 5 x^2-2/5x+3/5=0--eq1 Compare eq1 above with x^2-(alpha+Beta)x+alphabeta=o alpha+beta=2/5 alphabeta=3/5 11ai) alpha^3+beta^3=(alpha+beta)(alpha+beta)^2-3alphabeta =2/5(2/5)^2-3(3/5) =2/5(4/25-9/5) =2/5(-41/25) =-82/125 11aii) Sum of alpha +1/beta and beta+1/alpha =(alphabeta+1)/beta +(alpha beta + 1)/alpha =alphabeta^2+alpha^2beta+beta/alphabeta =alphabeta(alpha+beta)+(alpha+beta)/alphabeta product root (alpha +1/beta)(beta+1/alpha) =alpha(beta+1/alpha)=1/beta(beta+1/alpha) =alphabeta+1+1+1q/alphabeta sum=alphabeta(alpha+beta)+(alpha+beta)/alphabeta =3/5(2/5)+(2/5) =6/25+2/5 =(6+10)/25 =16/25 Product alphabeta+2+1/alphabeta =3/5+2+1/(3/5) =3/5+2/1+5/3 =64/15 Recall x^2-(alpha+beta)+alphabeta=0 x^2-16/25x+64/15=0 11bi) f(x)=x^3+2x-kx-6 x-2=0 x=+2 f(2)=2^3+2(2)^2-k(2)-6=0 8+8-2k-6=0 16-6-2k0 10-2k=0 2k=10 k=10/2 k=5 11bii) (x^2+4x+3) (x-2)rootx^3+2x^2-5x-6 =x^3-2x^2 =4x^2-5x =4x^2-8x =3x-6 =3x-6 =0 =>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)
11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3
10a)
T4=ar^4=162---(1)
T8=ar^7=4374---(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6---(1)
T4=ar^3=54---(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309
15b)
Pr
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1)
A=(2/1 -1/3)
|A|=2*3-1x-1
=6+1=7
minor(2)=3
(-1)=1
(1)=-1
(3)=2
minor A=(3/-1 1/2)
co-factor A(3/1 -1/2)
adjoint (A)=(3/-1 1/2)
A^-1=1/|A|adj(A)
=1/7(3/-1 1/2)
2)
3)
4)
10)
1b) Now Posted Click Image Below To View
5)
diagram
final velocity(v)=u +at
=20+5(4)
=20+20
=40m/s total distance= area of trapezium OABD
+Area of a triagngle BCD
time between DC=t
from v=u+at
t=v-u/a= 40-0/10= 4s
total distance= 1/2(20+4)4 +1/2(4) * 40=
1/2(60)4 + 1/2(4) *40
=30*4+ 2*40
=120+80
=200m
5b)
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10a)Click Image Below To View
10b)
n/2 (b+a)=330
n/2 (60+6)=330
66n/66=660/66
n=10
The number of terms = 10
10c)
ar^3 = 54.......(i)
ar = 6.......(ii)
√r^2 =3
Substitute for the value or r in eqn (ii)
3a/3 = 6/3
a=2
Tn = 2*3^n-1
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15b)
Pr(at most 2)=Pr(0)+Pr(1)+Pr(2)
n=1000,Pr=0.05 Pr(2)=0.95
Pr(x)=e^-landa(landa^x)
Pr(0.05)<0.1 and landa=1.000*0.05=50 e^-50(50^0)/o!+e^-50(50^1)/1!+e^-50(50^2)/2! 1/e^50+50/e^50+250.2e^50 =1/e^50[1+50+25/2] ----------------- 12a) w=weight of the plank moment=force*perpendicular distance clockwise moment=Anticlockwise Make A the reference point Pw=2.5m, AM=(2.5-0.8)m =1.7m g=10m/s AB=3.2m,PA=0.8m Convert 30kg to weight =mg=30*10=300N therefore 300*0.8=w*1.7 w=240/1.7=141.18N Distance of its C.G from P=2.5m 12b) For R1=A, R2=B R1=R2=2R1 Let Q be the reference point 300x*141.18*2.5=2R1(4.8+1) 300x+352.95=11.6R1 300x-11.6R1=-352.95---(1) Let P in the reference point 2R1(0.8+4)=141.15*2.5 9.6R1=352.95 R1=352.95/9.6 =36.77N 1a) A=(2 -1)(1 3) |A|=(2*3--1) |A|=6+1 |A|=7 since the determinant of A is not 0 the A is not non singular matrix To find A^-1 A^-1=(3 1)(-1 2) then 1/7(3 1)(-1 2) A^-1=(3/7 1/7)(-1/7 2/7) 2a) 3root2-3root5/5root2+2roo5 conjugate=5root2-2root5 =(3root2-3root5)(5root2-2root5)/(5root2+2root5)(5root2-2root5) =(15*2-6root10-15root10-6*5)/(25*2-10root10+10root10-4*5) =(30-31root10-30)/(50-20) =-31root10/30 2b) cosx=adj/hyp=12/13 then tanx=opp/adj=5/12 siny=3/5 tany=3/4 tan(x+y)=tanx+tany/1-tanxtany =(5/12+3/4)/(1-(5/12)(3/4) =(14/12)/(1-15/48) =36/125+18/125+16/125 =70/125 =14/25 3a) Pr(A passed)=4/5,pr(A failed)=1/5 Pr(B passed)=3/5,Pr(B failed)=2/5 Pr(C passed)-2/5,Pr(C failed)=3/5 Pr(atleast one passed) =4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5 =24/125+9/125+4/125 =47/125 3b) Pr(atleast one failed) =4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5 =101/125 4) (3+2y)^5 with (a+b)^5 shows a=3,b=2y Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5 and 1 (3+2y)^5 =(3)^5+5(3)^4(2y)+10(3)^3(2y)^2+10(3)^2(2y)^3+5(3)(2y)^4+(2y)^5 =243+810y+1080y^2+720y^3+240y^4+32y^5 (3.04)^5=(3+0.04)^5 =3^5+5(3)^4(0.04)+10(3)^2(0.04)^2+10(3)^2(0.04)^3+5(3)(0.04)^4+(0.04)^5 =243+5*81(0.04)+10*27(0.0016)+10*9(0.000064)+15(0.0000026)+0.0000001 =243+16.2+0.432+0.00576+0.000039+0.0000001 =259.63783 11a) 5x^2-2x+3=0 Divide both sides by 5 x^2-2/5x+3/5=0--eq1 Compare eq1 above with x^2-(alpha+Beta)x+alphabeta=o alpha+beta=2/5 alphabeta=3/5 11ai) alpha^3+beta^3=(alpha+beta)(alpha+beta)^2-3alphabeta =2/5(2/5)^2-3(3/5) =2/5(4/25-9/5) =2/5(-41/25) =-82/125 11aii) Sum of alpha +1/beta and beta+1/alpha =(alphabeta+1)/beta +(alpha beta + 1)/alpha =alphabeta^2+alpha^2beta+beta/alphabeta =alphabeta(alpha+beta)+(alpha+beta)/alphabeta product root (alpha +1/beta)(beta+1/alpha) =alpha(beta+1/alpha)=1/beta(beta+1/alpha) =alphabeta+1+1+1q/alphabeta sum=alphabeta(alpha+beta)+(alpha+beta)/alphabeta =3/5(2/5)+(2/5) =6/25+2/5 =(6+10)/25 =16/25 Product alphabeta+2+1/alphabeta =3/5+2+1/(3/5) =3/5+2/1+5/3 =64/15 Recall x^2-(alpha+beta)+alphabeta=0 x^2-16/25x+64/15=0 11bi) f(x)=x^3+2x-kx-6 x-2=0 x=+2 f(2)=2^3+2(2)^2-k(2)-6=0 8+8-2k-6=0 16-6-2k0 10-2k=0 2k=10 k=10/2 k=5 11bii) (x^2+4x+3) (x-2)rootx^3+2x^2-5x-6 =x^3-2x^2 =4x^2-5x =4x^2-8x =3x-6 =3x-6 =0 =>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)
11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3
10a)
T4=ar^4=162---(1)
T8=ar^7=4374---(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6---(1)
T4=ar^3=54---(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309
15b)
Pr
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1-10: DCEBDBABAC.
11-20: ABDDDCCDAC.
21- 30: CDBACAEBCB.
31-40: CEAEEDEBBD
41-50: CBBADDCDDC
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Choose From Number 1,2,3,3,4 ,7,8,9.
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1)
|A|=2*3-1x-1
=6+1=7
minor(2)=3
(-1)=1
(1)=-1
(3)=2
minor A=(3/-1 1/2)
co-factor A(3/1 -1/2)
adjoint (A)=(3/-1 1/2)
A^-1=1/|A|adj(A)
=1/7(3/-1 1/2)
2)
3)
4)
10)
1b) Now Posted Click Image Below To View
5)
diagram
final velocity(v)=u +at
=20+5(4)
=20+20
=40m/s total distance= area of trapezium OABD
+Area of a triagngle BCD
time between DC=t
from v=u+at
t=v-u/a= 40-0/10= 4s
total distance= 1/2(20+4)4 +1/2(4) * 40=
1/2(60)4 + 1/2(4) *40
=30*4+ 2*40
=120+80
=200m
5b)
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10a)Click Image Below To View
10b)
n/2 (b+a)=330
n/2 (60+6)=330
66n/66=660/66
n=10
The number of terms = 10
10c)
ar^3 = 54.......(i)
ar = 6.......(ii)
√r^2 =3
Substitute for the value or r in eqn (ii)
3a/3 = 6/3
a=2
Tn = 2*3^n-1
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