Answer:
Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is 
The probability mass function for the random variable is given by:
The best answer for this case would be:
C. Poisson distribution
Step-by-step explanation:
Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is
The probability mass function for the random variable is given by:
And f(x)=0 for other case.
For this distribution the expected value is the same parameter 
And for this case we want to calculate this probability:
The best answer for this case would be:
C. Poisson distribution
Solving the complex number, we get the value of the missing value that is ‘a’ = -6
We have been given the expression as
|a – i| = √37 (1)
Which is an expression of complex number. The general expression of complex number is given as
z = x + iy
where x is the real part and iy is the imaginary part
To find the modulus value, the formula is given by,
|z| = |x + iy|
|z| = √[(real part)2 + (imaginary part)2]
|z| = √(x2 + y2)
According to the question, |z| = √37 (2)
Equating equation (1) and (2), we get
√(a2 + 1) = √37
(a2 + 1) = 37
a2 = 37 – 1
a2 = 36
a = √36
a = ±6
Now value of a can be 6 or -6. We have been given that the modulus is in third quadrant.
Hence the value will be negative. Therefore, the missing value will be -6.
Learn more about complex number here : brainly.com/question/5564133
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3.10
I think, Im very sorry if this is wrong!
Answer:
G. 7
Explanation:
The longest side must be less than the sum of the two shorter sides.
The difference between the longest and shortest sides must be less than the middle side.
l+a = 11
l < (a + 4) ... l < (11 - l + 4)
... 2 l < 15 ... l < 7.5
--Variables
l = 7 (Longest Side/Answer)
a = 4 (Other Missing Length)
m = 4 (Side Given)
Answer:
The answer is "
"
Step-by-step explanation:
Given:
Find critical points:
differentiate the value with respect of x:
![\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]](https://tex.z-dn.net/?f=%5Cto%20g%27%28x%29%3D%20%28x-e%29%5E3%20%5Cfrac%7Bd%7D%7Bdx%7De%5E%7Be-r%7D%20%2Be%5E%7Be-r%7D%20%20%5Cfrac%7Bd%7D%7Bdx%7D%28x-e%29%5E3%3D%28x-e%29%5E2%20e%5E%7B%28e-x%29%7D%20%5B-x%2Be%2B3%5D)
critical points 
![\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3](https://tex.z-dn.net/?f=%5Cto%20%28x-e%29%5E2%20e%5E%7B%28e-x%29%7D%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20e%5E%7B%28e-x%29%7D%5Cneq%200%20%5C%5C%5C%5C%5Cto%20%28x-e%29%5E2%3D0%5C%5C%5C%5C%20%5Cto%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20x%3De%5C%5C%5C%5C%5Cto%20x%3De%2B3%5C%5C%5C%5C%5Cto%20x%3D%20e%2Ce%2B3)
So,
The critical points of 
