Answer:
x = StartFraction negative
(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction
Step-by-step explanation:
0 = – 3x2 – 2x + 6
It can still be written as
– 3x2 – 2x + 6 =0
Quadratic formula=
-b+or-√b^2-4ac/2a
Where
a=-3
b=-2
c=6
x= -(-2)+ or-√(-2)^2-4(-3)(6)/2(-3)
x = StartFraction negative
(negative 2) plus or minus StartRoot (negative 2) squared minus 4 (negative 3)(6) EndRoot Over 2(negative 3) EndFraction
Answer:
E(X) = 1.28
Var(X) = 0.6016
E(X | Y=2) = 1.6667
Var(X | Y=2) = 0.4272
Step-by-step explanation:
Hello,
Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.
Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)
As sin x *cos x=sin (2x) /2
P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
*1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a
sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)
So sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".
Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)
4 tables because the total amount of people is 24 and if you divide by 6 you get 4 so the greatest amount of tables is 4. There will be 6 adults and 1 kid at each table
Hey next time use your calculator, if you can.
Since these problems are using this equation :
we can derive into this immediately.
Q1: 525 (Try it, don't copy, it's not nice ;) ).
Q2: 865.28
Q3: By using logs: 0.0566 years. (I'm not sure on that one...) Sry