What is the slop-intercept of the line that passes through the points (7,-5) and (3,-9)?

1 answer:

4 0

$\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-9}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-(-5)}{3-7}\implies \cfrac{-9+5}{-4}\implies \cfrac{-4}{-4}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-5)=1(x-7) \\\\\\ y+5=x-7\implies y=x-12$

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The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

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71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: