What would be a equation for (5,2) and (0,7)?

1 answer:

6 0

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{7}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{5}}}\implies \cfrac{5}{-5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-1}(x-\stackrel{x_1}{5}) \\\\\\ y-2=-x+5\implies y = -x+7$

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you are rolling a 10-sided die with sides numbered 1-10. you keep rolling the die until you roll a prime number and then stop. l

evablogger [386]

The formula for E(x) would be SUM[ X(n)P(n)] from n = 1 to n = infinity until we get the prime number

In the given problem, a 10-sided die with 1-10 numbers is rolled until we get a prime number

Prime numbers between 1-10 = 2,3,5,7 (4 prime numbers)

Non - prime numbers = 1,4,6,8,9,10( 6 non-prime number)

Let X be the number of times the die is rolled

The probability of getting a prime =

$P(Prime) = \frac{4}{10}$

Now, the value of

E(X)=∑x.P(x) [ x-> {1, infinity}]

= P(1)X(1) +P(2)X(2)+P(3)X(3)+P(4)X(4)........+ P(n)X(n)

= 1. $\frac{4}{10}$ + 2. $\frac{6}{10}$ . $\frac{4}{10}$ + 3. $\frac{6}{10}$ . $\frac{6}{10}$ $\frac{4}{10}$ + 4. $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{6}{10}$ . $\frac{4}{10}$ .......