$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-2-(-1)}{6-3}\implies \cfrac{-2+1}{3}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-1)=-\cfrac{1}{3}(x-3)\implies y+1=-\cfrac{1}{3}(x-3)$

8 0

4 years ago

A company uses three different assembly lines- A1, A2, and A3- to manufacture a particular component. Of thosemanufactured by li

hammer [34]

Answer:

The probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

Step-by-step explanation:

The three different assembly lines are: A₁, A₂ and A₃.

Denote R as the event that a component needs rework.

It is given that:

$P (R|A_{1})=0.05\\P (R|A_{2})=0.08\\P (R|A_{3})=0.10\\P (A_{1})=0.50\\P (A_{2})=0.30\\P (A_{3})=0.20$

Compute the probability that a randomly selected component needs rework as follows:

$P(R)=P(R|A_{1})P(A_{1})+P(R|A_{2})P(A_{2})+P(R|A_{3})P(A_{3})\\=(0.05\times0.50)+(0.08\times0.30)+(0.10\times0.20)\\=0.069$

Compute the probability that a randomly selected component needs rework when it came from line A₁ as follows:

$P (A_{1}|R)=\frac{P(R|A_{1})P(A_{1})}{P(R)}=\frac{0.05\times0.50}{0.069} =0.3623$

Thus, the probability that a randomly selected component needs rework when it came from line A₁ is 0.3623.

6 0

3 years ago

The table below shows the outputs y for different inputs x:

igor_vitrenko [27]

Part A)
The table doesn't represent y as a function of x.
Why? One reason is that 1 or 3 is the first element in more than one ordered pair.
Part B)
f(x) = 10x + 20
f(120)= 10*120 + 20 = 1200 + 20 = 1220
f(120) represents cost of boating for 120 hours.

3 0

3 years ago

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