Solve by completing the square. 3x^2+x-2=0

Suppose the standard deviation of X is 6 and the standard deviation of Y is 8. Answer the following two questions, rounding to t

Darina [25.2K]

Recall that

Var[aX + bY] = a ² Var[X] + 2ab Cov[X, Y] + b ² Var[Y]

Then

Var[3X - 7Y] = 9 Var[X] - 42 Cov[X, Y] + 49 Var[Y]

Now, standard deviation = square root of variance, so

Var[3X - 7Y] = 9×6² - 42×2 + 49×8² = 3376

The general result is easy to prove: by definition,

Var[X] = E[(X - E[X])²] = E[X ²] - E[X]²

Cov[X, Y] = E[(X - E[X]) (Y - E[Y])] = E[XY] - E[X] E[Y]

Then

Var[aX + bY] = E[((aX + bY) - E[aX + bY])²]

… = E[(aX + bY)²] - E[aX + bY]²

… = E[a ² X ² + 2abXY + b ² Y ²] - (a E[X] + b E[Y])²

… = E[a ² X ² + 2abXY + b ² Y ²] - (a ² E[X]² + 2 ab E[X] E[Y] + b ² E[Y]²)

… = a ² E[X ²] + 2ab E[XY] + b ² E[Y ²] - a ² E[X]² - 2 ab E[X] E[Y] - b ² E[Y]²

… = a ² (E[X ²] - E[X]²) + 2ab (E[XY] - E[X] E[Y]) + b ² (E[Y ²] - E[Y]²)

… = a ² Var[X] + 2ab Cov[X, Y] + b ² Var[Y]

6 0

3 years ago

Simplify (20^13)^19 A. 400^32 B. 20^32 C. 20^247 D. 400^247

vovikov84 [41]

When an exponent is outside the parenthesis, you multiply them.

13*19=247

“C” 2^247 is the answer.

I hope this helps!
~kaikers

5 0

4 years ago

the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,

kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>Binomial probability distribution </h3>

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$