Question

sum

Answer 1

\[\sum_{k=2}^{6}(2i-3) \]

Answer 2

1 + 4 + 7 + 10 + 13
2 + 4 + 6 + 8 + 10
-1 + 1 + 3 + 5 + 7
1 + 3 + 5 + 7 + 9

Answer 3

@Michele_Laino

Answer 4

Which series corresponds to this summation?

Answer 5

Hint:
I think that your sum is:
\[\begin{gathered}
\sum\limits_{k = 2}^6 {\left( {2k - 3} \right)} = \hfill \\
\hfill \\
= \left( {2 \times 2 - 3} \right) + \left( {2 \times 3 - 3} \right) + \left( {2 \times 4 - 3} \right) + \left( {2 \times 5 - 3} \right) + \left( {2 \times 6 - 3} \right) = ...? \hfill \\
\end{gathered} \]

Answer 6

1

Answer 7

3

Answer 8

5

Answer 9

you know about probability distribution

Answer 10

we have:
\[\begin{gathered}
\sum\limits_{k = 2}^6 {\left( {2k - 3} \right)} = \hfill \\
\hfill \\
= \left( {2 \times 2 - 3} \right) + \left( {2 \times 3 - 3} \right) + \left( {2 \times 4 - 3} \right) + \left( {2 \times 5 - 3} \right) + \left( {2 \times 6 - 3} \right) = \hfill \\
= 1 + 3 + 5 + 7 + 9 = ...? \hfill \\
\end{gathered} \]

Answer 11

P(xi)(xi −

Answer 12

\[P(x i)(x i-\mu)^2\]

Answer 13

that formula

Answer 14

I need your help with one other question

Answer 15

ok here it is

Answer 16

The probability distribution of X is given in the table,