Answer:
2Ab^2
Step-by-step explanation:
I think its right
Answer:
3 x^3 y^4 sqrt(5x)
Step-by-step explanation:
sqrt(45x^7y^8)
We know that sqrt(ab) = sqrt(a)sqrt(b)
sqrt(45)sqrt(x^7) sqrt(y^8)
sqrt(9*5) sqrt(x^2 *x^2 * x^2* x) sqrt(y^2 *y^2 *y^2 *y^2)
We know that sqrt(ab) = sqrt(a)sqrt(b)
sqrt(9)sqrt(5) sqrt(x^2)sqrt(x^2) sqrt(x^2) sqrt(x) sqrt(y^2)sqrt(y^2)sqrt(y^2)sqrt(y^2)
3 sqrt(5) x*x*x sqrt(x) y*y*y*y
3 x^3 y^4 sqrt(5)sqrt(x)
3 x^3 y^4 sqrt(5x)
Step-by-step explanation:
( y = 4x/5 + 2)
m1.m2 = -1
4/5.m2 = -1
m2 = -5/4
formula = y - y1 = m2 ( x - x1)
y - 3 = -5/4 ( x - 6)
y -3 = -5x /4 + 30/4
y = -5x /4 + 21/2
Answer:
the approximate probability that the insurance company will have claims exceeding the premiums collected is 
Step-by-step explanation:
The probability of the density function of the total claim amount for the health insurance policy is given as :

Thus, the expected total claim amount
= 1000
The variance of the total claim amount 
However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100
To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :
P(X > 1100 n )
where n = numbers of premium sold





Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is 
Order doesn't mater so use combinations:
6C3 = 6!/ (3! (6-3)! ) = 6! /(3!*3!) = 6*5*4*3! / (3! 3!) = 6*5*4/3*2*1) = 20