Answer: Mean = 2.91 and Variance = 0.087
Step-by-step explanation:
Given : The probability of a correct classification of any part p= 0.97.
The number of parts are inspected : n=3
Let the random variable X denote the number of parts that are correctly classified.
For binomial experiment , the mean and variance is given by :-
Then, the the mean and variance of X will be :-
![\mu=3(0.97)=2.91\\\\\sigma^2=np(1-p)=3(0.97)(0.03)=0.0873\approx0.087\ \ [ \text{Rounded to three decimal places}]](https://tex.z-dn.net/?f=%5Cmu%3D3%280.97%29%3D2.91%5C%5C%5C%5C%5Csigma%5E2%3Dnp%281-p%29%3D3%280.97%29%280.03%29%3D0.0873%5Capprox0.087%5C%20%5C%20%5B%20%5Ctext%7BRounded%20to%20three%20decimal%20places%7D%5D)
Answer:
-1 -+i/5/2
Step-by-step explanation:
No soy una calculadora y son las 12am
Answer:
x = -2, y = -2
Step-by-step explanation:
Your goal is to try and cancel out a variable. I want to get rid of y so i subtracted the first equation from the second one.
After that I solve for x and got x=-2.
I used x=-2 and plugged it back into either of the equation to solve for y.
Step-by-step explanation:
t=2+5×j
j=5
t=2+5×5
t=27
