Answer:
Use the formula π*(r^2) where r is radius
Area of big circle, all 3=314.1592654 (approximately) and this is =100%
Area of middle circle=153.93804
Area of small circle=78.53981634
Percentage of middle circle with small circle = (153.93804/314.1592654)*100= 48.999999999999 approx= 49%
Percentage of small circle alone = (78.53981634/314.1592654)*100
= 25%
So 51%= big circle alone
And 51%+25%= 76%
100%-76%=24%
Hi there!
5k - 2k = 12
Solve for K
3k = 12
Divide both sides by 3
3k/3 = 12/3
k = 4
The correct answer is : 4
I hope that helps!
Brainliest answer :)
Create the sum of the given polynomial terms.
f(x,y) = 3x² - 2x²y + 2xy² + 4y² + xy² - y² + 2x²y
= 3x² + 3xy² + 3y²
Test the given terms to see if they belong in f(x,y).
3: NO
3x²: YES
3y²: YES
3xy²: YES
4x²: NO
Answer:
The terms in the sum of the given polynomials are 3x², 3y², and 3xy².
Answer:

In order to find the variance we need to find first the second moment given by:

And replacing we got:

The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:

Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:

And replacing we got:

In order to find the variance we need to find first the second moment given by:

And replacing we got:

The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:

It is
12000 x (1.06)^12 + 50000 x (1.06)^6
= 95,072.31
start of 4th year to end of 6th year = 6 semi-annual periods where interest is compounded for the second deposit