First question seems incomplete :
Answer:
40 ways
Step-by-step explanation:
Question B:
Number of boys = 6
Number of girls = 4
Number of people in committee = 3
Number of ways of selecting committee with atleast 2 girls :
We either have :
(2 girls 1 boy) or (3girls 0 boy)
(4C2 * 6C1) + (4C3 * 6C0)
nCr = n! ÷ (n-r)!r!
4C2 = 4! ÷ 2!2! = 6
6C1 = 6! ÷ 5!1! = 6
4C3 = 4! ÷ 1!3! = 4
6C0 = 6! ÷ 6!0! = 1
(6 * 6) + (4 * 1)
36 + 4
= 40 ways
Step-by-step explanation:
3a^2+24a+45=0
×3/3( first ×3)
9a^2+24(3a)+135=0
(3a+15)(3a+9)=0
now /3
3(a+5)(a+3)
a=-5✓
a=-3✓
Answer:
Bias for the estimator = -0.56
Mean Square Error for the estimator = 6.6311
Step-by-step explanation:
Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.
To find - Determine the bias and the mean squared error for this estimator of the mean.
Proof -
Let us denote
X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)
Now,
An estimate of mean, μ is suggested as

Now
Bias for the estimator = E(μ bar) - μ
= 
= 
= 
= 
= 
= 3.9375 - 4.5
= - 0.5625 ≈ -0.56
∴ we get
Bias for the estimator = -0.56
Now,
Mean Square Error for the estimator = E[(μ bar - μ)²]
= Var(μ bar) + [Bias(μ bar, μ)]²
= 
= 
= ![\frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})] }) + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%28%20%5B%7B3Var%28X_%7B1%7D%29%20%2B%204Var%28X_%7B2%7D%29%5D%20%20%7D%29%20%2B%200.3136)
= ![\frac{1}{64} [{3(57.76) + 4(57.76)}] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B%7B3%2857.76%29%20%2B%204%2857.76%29%7D%5D%20%20%7D%20%2B%200.3136)
= ![\frac{1}{64} [7(57.76)}] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B7%2857.76%29%7D%5D%20%20%7D%20%2B%200.3136)
= ![\frac{1}{64} [404.32] } + 0.3136](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B64%7D%20%5B404.32%5D%20%20%7D%20%2B%200.3136)
= 
= 6.6311
∴ we get
Mean Square Error for the estimator = 6.6311
There is no graph to support the statement and so, this invisible graph is misleading :)