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statuscvo [17]
3 years ago
13

The exponential probability distribution is a discrete distribution that is often used to describe time between customer arrival

s.
Mathematics
1 answer:
n200080 [17]3 years ago
4 0

Answer:

True

Step-by-step explanation:

The time between customer arrivals is called inter-arrival time. According to Queueing Notation, the inter-arrival time can be model based on difference probability distribution. The probability distribution by which the inter-arrival time can be modeled include:

  • Exponential Distribution or Markov distribution
  • Constant or Deterministic
  • Hyper - exponential
  • Arbitrary or General distribution
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Which of the following equations is equivalent to x - y = 6?
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I believe the answer would be A. 2x-y=6 :)
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3 years ago
All vectors are in Rn. Check the true statements below:
Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that ||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}. Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then ||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||.

C) Consider S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2. This set is orthogonal because (0,2)\cdot(2,0)=0(2)+2(0)=0, but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in \mathbb{R}^n. Then the columns of A form an orthonormal set. We have that A^{-1}=A^t. To see this, note than the component b_{ij} of the product A^t A is the dot product of the i-th row of A^t and the jth row of A. But the i-th row of A^t is equal to the i-th column of A. If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then A^t A=I    

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set \{u_1,u_2\cdots u_p\} and suppose that there are coefficients a_i such that a_1u_1+a_2u_2\cdots a_nu_n=0. For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then a_i||u_i||=0 then a_i=0.  

5 0
3 years ago
Students at Mendel Middle School are planning a fair for their school's fundraiser. Liam volunteers to help plan the event. The
FinnZ [79.3K]

Answer:

15 hours

Step-by-step explanation:

x is the number of hours that Liam volunteered in the first month

The next month he volunteered 1\dfrac{2}{3} times the time he volunteered the first month.

Total time he volunteered in the two months is 40 hours so

x+1\dfrac{2}{3}x=40\\\Rightarrow x+\dfrac{5}{3}x=40\\\Rightarrow \dfrac{3x+5x}{3}=40\\\Rightarrow x=\dfrac{40\times 3}{8}\\\Rightarrow x=15

Time he volunteered in the first month is 15 hours.

3 0
2 years ago
Compute the conditional probabilities from the two-way frequency table.
Natali5045456 [20]

Answer:

The table is show clearly in the figure attached.

P(boy if favourite activity is swimming) = 8/17 = 0.47

P(girl if favourite activity is sport) = 7/27 = 0.26

P(girl if favourite activity is reading) = 4/6 = 0.67

P(boy if favourite activity is sport) = 20/27 = 0.74

P(favourite activity is swimming if a girl) = 9/20 = 0.45

P(favourite activity is reading if a boy) = 2/30 = 0.07

P(favourite activity is swimming if a boy) = 8/30 = 0.27

P(favourite activity is reading if a girl) = 4/20 = 0.2

6 0
3 years ago
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