Answer:
[7 38 9] is the answer
Step-by-step explanation:
Hope this helps I did my calculations and this is the answer
Answer:
0.9 ≈ 90%
Step-by-step explanation:
60% = 60/100
150% of 60% = 150/100 * 60/100
= 9000/10000
= 0.9 ≈ 90%
I think you have to first separate the integral:1/(1+v^2) + v/(1+v^2),
so the integral of the first term is ArcTan (v) and for the integral of the second term i recommend you to do a change of variable:
y= 1+v^2
so
dy= 2v
and
v= dy/2and then you substitute:v/(1+v^2) = (1/2)(dy/y)
and the integral is
(1/2) (In y)finally you plug in the initial variables:
(1/2)(In [1+v^2])
so the total integral is:
ArcTan (y) + (1/2)(In [1+v^2])
In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function f applied to an input x gives a result of y, then applying its inverse function g to y gives the result x, and vice versa, i.e., f(x) = y if and only if g(y) = x. <u><em>Courtesy to Wikipedia</em></u>
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560



P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = 
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)


From z tables
P(Y > 20)
0.0903