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Katarina [22]
3 years ago
11

The equation of a line that goes through the points (0,0) and (300000,365)

Mathematics
1 answer:
olga_2 [115]3 years ago
4 0
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 0 &,& 0~) 
%  (c,d)
&&(~ 300000 &,& 365~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{365-0}{300000-0}\\\\\\ \cfrac{365}{300000}\implies \cfrac{73}{60000}

\bf \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=\cfrac{73}{60000}(x-0)\implies y=\cfrac{73}{60000}x
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Answer:

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And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

Step-by-step explanation:

For this case we have the following distribution given:

X        0         1       2       3       4         5        6

P(X)  0.3   0.25   0.2   0.12   0.07   0.04   0.02

For this case we need to find first the expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And replacing we got:

E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61

Now we can find the second moment given by:

E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97

And the variance would be given by:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

 

8 0
3 years ago
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Answer:

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