(7b)
m2u2-m1u1=V(m1+m2)
6*5-3*8=V(3+6)
30-24=9v
9v=6
v=6/9
v=0.67m/s
—————-
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
====================
.
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
.
12ii)
CLICK HERE TO VIEW THE IMAGE
==================================
(13ai)
M=2
P=5
C=3
total=10
If the books of the same subject are to stand together
No of arrangements=2!*5!*3!*3!
=2*120*6*6
=8640arrangements
.
(13aii)
Only the physics textbook must stand together
No of arrangements=5!*6!
=120*720
=86400arrangements
.
(13b)
P=13/20
q=1-13/20=7/20
pr(atleast 3 speak E)=1-Pr(2 speak E)
=(1-8C1p^1q^7+8C2p^2q^6)
=1-(8*(13/20)*(7/20)^7+28(13/20)^2*(7/20)^6
=1-(0.003346+0.0217467)
=1-0.0251
=0.9749
=0.975(3s.f)
==================================
