Var[2X + 3Y] = 2² Var[X] + 2 Cov[X, Y] + 3² Var[Y]
but since X and Y are given to be independent, the covariance term vanishes and you're left with
Var[2X + 3Y] = 4 Var[X] + 9 Var[Y]
X follows an exponential distribution with parameter <em>λ</em> = 1/6, so its mean is 1/<em>λ</em> = 6 and its variance is 1/<em>λ</em>² = 36.
Y is uniformly distributed over [<em>a</em>, <em>b</em>] = [4, 10], so its mean is (<em>a</em> + <em>b</em>)/2 = 7 and its variance is (<em>b</em> - <em>a</em>)²/12 = 3.
So you have
Var[2X + 3Y] = 4 × 36 + 9 × 3 = 171
Answer:
392
Step-by-step explanation:
Total no = 2n²
= 2X14²
= 392
Answer:
Step-by-step explanation:
In general, a solution of a system in two variables is an ordered pair that makes BOTH equations true. In other words, it is where the two graphs intersect, what they have in common. So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system.
I just finished talking about ratios a few days ago at school, the answer is 4 cups of pretzels : 2 cups of bread chips