Answer:
x < -3
Step-by-step explanation:
Step 1: Subtract 6 from both sides.
Step 2: Divide both sides by 4.
Therefore, the answer is x < -3.
Answer: -8-10
Step-by-step explanation:
Answer:
See the proof below.
Step-by-step explanation:
Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."
Solution to the problem
For this case we can assume that we have N independent variables
with the following distribution:
bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:
From the info given we know that
We need to proof that
by the definition of binomial random variable then we need to show that:


The deduction is based on the definition of independent random variables, we can do this:

And for the variance of Z we can do this:
![Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2](https://tex.z-dn.net/?f=%20Var%28Z%29_%20%3D%20E%28N%29%20Var%28X%29%20%2B%20Var%20%28N%29%20%5BE%28X%29%5D%5E2%20)
![Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5Bp%281-p%29%5D%20%2B%20Mq%281-q%29%20p%5E2)
And if we take common factor
we got:
![Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]](https://tex.z-dn.net/?f=%20Var%28Z%29%20%3DMpq%20%5B%281-p%29%20%2B%20%281-q%29p%5D%3D%20Mpq%5B1-p%20%2Bp-pq%5D%3D%20Mpq%5B1-pq%5D)
And as we can see then we can conclude that 
Answer:
x = -2, y=1
Step-by-step explanation:
x-7y=-9---------------------equation 1
-x+8y=10-------------------equation 2
From equation 1, make x the subject of formula
x=7y-9----------------------equation 3
substituting x=7y-9 in equation 2,
-(7y-9)+8y=10
Expanding bracket
-7y+9+8y=10
Collecting like terms
8y-7y=10-9
y=1
substituting y=1 in 3
x=7(1)-9
x=7-9
x=-2
Answer:
v=3
u=6
x=2√2
y=2√2
Step-by-step explanation:
First triangle :
tanΦ=v/3√3
tan30=v/3√3
(√3)/3=v/(3√3)
Cross multiply
(√3)/3 x 3√3=v
(√3 x 3√3)/3=v
(3√9)/3=v
(3x3)/3=v
9/3=v
3=v
v=3
Sin30=v/u
0.5=3/u
Cross multiply
0.5xu=3
0.5u=3
Divide both sides by 0.5
0.5u/0.5=3/0.5
u=6
Second triangle :
sin45=x/4
Cross multiply
x=4 x sin45
x=4 x (√2)/2
x=2√2
Cos45=y/4
Cross multiply
4 x Cos45=y
4 x (√2)/2=y
(4√2)/2=y
2√2=y
y=2√2