How do i work out the following simultaneous equations?
x+3y=11 and 2x-y=11
you, this is part of my half term homework and i really need to understand how to do this as soon as possible , you!

(a) The mean is 0.04

The standard deviation is approximately 0.0196

(b) The sampling distribution of p is not approximately normal because n·p is less than 10

(c) Given that n  = 400, we have;

i) The mean is 0.04

The standard deviation is approximately 0.0098

ii) The sampling distribution is always centered at the population mean, regardless of the sample size

iii) When the sample size increases, the standard deviation decreases

(d) The sampling distribution of p is approximately normal because n·p and n·(1 - p) are both at least 10

Step-by-step explanation:

(a) From the question, we have that the actual proportion of all adult Americans who watch online = 0.04

Given that the sample size, 'n' is 100 which is larger than 30, we can assume that the distribution of the sample mean is normal

Therefore, the mean (proportion) of the sample, P = The mean (proportion) of the population = 0.04

The mean, P = 0.04

The standard deviation, σ = √((p·(1 - p))/n)

∴ σ = √((0.04·(1 - 0.04))/100) ≈ 0.0196

(b) The sampling of p is approximately normal, given that the sample size, 'n' = 100 which is randomly selected from the population, we have;

The sampling distribution = n·p  = 100 × 0.04 = 4 <10

Therefore, the sample is not approximately normal because the sampling distribution, n·p = 4 which is less than 10

c) Given that n  = 400, we have;

p = 0.04 = The [population mean

The standard deviation, σ √((0.04·(1 - 0.04))/400) ≈ 0.0098

According to the central limit theorem when the sample size increases, the mean approaches the population mean, therefore, the sampling distribution is always centered at the population mean, regardless of the sample size

When the sample size increases, the standard deviation decreases

(d) Given that when the sample size, n = 400, we have;

n·p = 400 × 0.04 = 16 > 10

Similarly, we have;

n·(1 - p) = 400 × (1 - 0.04) = 384 > 10

The sample is approximately normal

Therefore, the sampling distribution of p is approximately normal because np and n(1 - p) are both at least 10