So far she scored 11 out of 15, which is a chance of 11/15, which is also the experimental probability for her next throw.
Answer:
Step-by-step explanation:
Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = points scored by students
u = mean time
s = standard deviation
From the information given,
u = 102 minutes
s = 18 minutes
1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as
P(x > 120 minutes) = 1 - P(x ≤ 120)
For x = 120
z = (120 - 102)/18 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
P(x > 120) = 1 - 0.8413 = 0.1587
2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as
P(x < 66 minutes)
For x = 66
z = (66 - 102)/18 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
P(x < 66 minutes) = 0.02275
Answer:
The fourth term of the expansion is -220 * x^9 * y^3
Step-by-step explanation:
Question:
Find the fourth term in (x-y)^12
Solution:
Notation: "n choose k", or combination of k objects from n objects,
C(n,k) = n! / ( k! (n-k)! )
For example, C(12,4) = 12! / (4! 8!) = 495
Using the binomial expansion formula
(a+b)^n
= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n
For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is
C(n,3)a^(n-3)b^3
=C(12,3) * x^(12-3) * (-y)^(3)
= 220*x^9*(-y)^3
= -220 * x^9 * y^3
The given equation is

This ODE (Ordinary Differential Equation) is separable.
That is,

Integrate to obtain

where k, c are constants.
Answer:

, c = constant.