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rjkz [21]
2 years ago
6

Which point is located at (-2, 5)? point C point D point B point A Image below

Mathematics
2 answers:
Firdavs [7]2 years ago
4 0

Answer:

B

Step-by-step explanation:

It is on the -2 5 so therefore your answer is B.

D is on 2 and -5
C is on 5 and -2
A is on -5 and 2

Therefore, it's B confirmed.

Svetllana [295]2 years ago
3 0

Answer:

point B

Step-by-step explanation:

point B is located at (-2, 5).

Hope it helps you in your learning process.

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Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

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4 years ago
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