parallel lines, have the same exact slope, hmmm what is the slope of y = -2/3 x + 1/3 anyway? well, low and behold, the equation is already in slope-intercept form, therefore
has a slope of -2/3.
so we're really looking for a line whose slope is -2/3 and runs through 9,4.
![\bf (\stackrel{x_1}{9}~,~\stackrel{y_1}{4})~\hspace{10em}slope = m\implies -\cfrac{2}{3}\\\\\\\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=-\cfrac{2}{3}(x-9)\\\\\\y-4=-\cfrac{2}{3}x+6\implies y=-\cfrac{2}{3}x+10](https://tex.z-dn.net/?f=%20%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B9%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29~%5Chspace%7B10em%7Dslope%20%3D%20%20m%5Cimplies%20-%5Ccfrac%7B2%7D%7B3%7D%5C%5C%5C%5C%5C%5C%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-4%3D-%5Ccfrac%7B2%7D%7B3%7D%28x-9%29%5C%5C%5C%5C%5C%5Cy-4%3D-%5Ccfrac%7B2%7D%7B3%7Dx%2B6%5Cimplies%20y%3D-%5Ccfrac%7B2%7D%7B3%7Dx%2B10%20)
THE ANSWER IS D whoops lol sorry i was in an argument lol
Answer:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=13.1](https://tex.z-dn.net/?f=%5Csigma%3D13.1)
We are interested on this probability
![P(X>140)](https://tex.z-dn.net/?f=P%28X%3E140%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
And we can find this probability using the complement rule:
![P(z>1.924)=1-P(z](https://tex.z-dn.net/?f=P%28z%3E1.924%29%3D1-P%28z%3C1.924%29%3D0.0272)