1.From the twelve people consisting of eight men and four women, will set up study groups of four people. If the group learned that there are least two men, then how many ways to form the study groups ?

2.Determine whether the following series convergent or divergent !
∑_(n=1)^∞▒n^2/n!

3.A mother has five children. Oldest child was aged 2p years, the youngest was aged p years. The three others orderly was aged 2p-2, p+2, and p+1 years. If they were average aged 17 years, so age the oldest child was ∙∙∙ years.

4.Find the solution of the initial value problem
dy/dx+y-1=t.e^(-t) ,y(0)=2

5. Statistical test value
Value Frequency
61 – 65 8
66 – 70 12
71 – 75 18
76 – 80 14
Determine the mode value of the data above!


Name : CORRY SORMIN
NIM : 08305141032

SOLVING:

1.From these problem, we known that twelve people consisting of eight men and four women, will set up group of four people which is contain at least two men. So, combination which possible to form are two men and two women, three men and woman, and only four men. We can solve it with combination formula. Many ways to form the study groups with at least two men:
= (two men, two women) + (three men, woman) + (four men)
= 8C2∙4C2 + 8C3∙4C1 + 8C4∙4C0
=8!/(8-2)!2! 4!/(4-2)!2!+8!/(8-3)!3! 4!/(4-1)!1!+8!/(8-4)!4! 4!/(4-0)!0!
=8!/6!2! 41/2!2!+8!/5!3! 4!/3!1!+8!/4!4! 4!/4!0!
=8.7.6!/6!2 4.3.2!/2!2+8.7.6.5!/5!3.2 4.3!/3!+8.7.6.5.4!/4!4.3.2 4!/4!0!
= 28 ∙ 6 + 56 ∙ 4 + 70 ∙ 1
= 168 + 224 + 70
= 462
So, many ways to form study groups were 462 ways.

2.By using the theorem in calculus, we obtain:
ρ=lim┬(n→∞)⁡〖〖(n+1)〗^2/(n+1)!〗 n!/n^2
ρ=lim┬(n→∞)〗⁡〖{〖(n+1)〗^2/(n+1)n!}.( n!/n^2) 〗
ρ=lim┬(n→∞)⁡〖〖(n+1)〗^2/((n+1).n^2 )〗
ρ=lim┬(n→∞)⁡〖(n+1)/n^2 〗
ρ= 0
So, we can conclude is a convergent series.

3.From the problem we known that were five children. Oldest child was 2p years, the youngest was aged p years and three others was aged 2p-2, p+2, and p+1 years. The average of five children were 17 years. So,
x ̅=(∑▒x_i )/n
x ̅=((2p)+(2p-2)+(p+2)+(p+1)+(p))/5
x ̅=(7p+1)/5
17= (7p+1)/5
84 = 7p
p = 12
So, age of the oldest child was 2p=2×12=24 years.

4.dy/dx+y-1=t.e^(-t)
dy/dx+y=t.e^(-t)+1 (linear differential equation)
Thus the integrating factor is
μ(t)=e^∫▒1dt
μ(t)=e^t
And on multiplying dy/dx+y=t.e^(-t)+1 by this quantity, we obtain:
e^t dy/dx+e^t y=e^t (te^(-t)+1)
e^t dy/dx+e^t y=t+e^t
the left side e^t dy/dx+e^t y=t+e^t is the derivative of ye^t. So, we can write this equation as (d(ye^t))/dt=t+e^t
And it follow by integration that
ye^t=∫▒(t+e^t )dt
ye^t=1/2 t^2+e^t+c, where c is an arbitrary constant. Therefore
y=(1/2 t^2+e^t+c)/e^t
y=1/2 t^2 e^(-t)+1+ce^(-t)
to satisfy the initial condition we substitute t=0 and y=2 and solve for c
2 = 1 + c
c = 1
we obtain the value c=1, so the solution of the given initial value problem is
y=1/2 t^2 e^(-t)+1+e^(-t)

5.Mode which is located on frequency of 18, we can using the formula
Mo=tb+(d_1/(d_1+d_2 ) )i
Tb = 70,5
d1 = 18 -12 = 6
d2 = 18 -14 = 4
i = 5
So,
Mo=70.5+6/(6+4) 5

Mo = 70,5 + 3
Mo = 73,5


By : ROCHMANIA SEPTIKASARI
NIM : 08305141005