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2 years ago

Solve each system by graphing.

Mathematics

1 answer:

Kryger [21]2 years ago

3 0

9514 1404 393

Answer:

(x, y) = (4, -4)

Step-by-step explanation:

A graphing calculator makes graphing very easy. The attachment shows the solution to be (x, y) = (4, -4).

The equations are in slope-intercept form, so it is convenient to start from the y-intercept and use the slope (rise/run) to find additional points on the line.

The first line can be drawn by staring at (0, -2) and moving down 1 grid unit for each 2 to the right.

The second line can be drawn by starting at (0, 2) and moving down 3 grid units for each 2 to the right.

The point of intersection of the lines, (4, -4), is the solution to the system of equations.

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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 <em>R</em> 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.

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2 years ago

A population of bacteria is initially 6000. After three hours the population is 3000. If this rate of decay continues, find the

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Answer:

The equation is $A=6000e^{-0.23t}$ at a rate of -23%.

Step-by-step explanation:

Decay can be represented by the equation $A=A_0e^{rt}$ . We can find the rate at which it decays by using t=3 hours and A=3000. This means $A_0=6000$ in this context.

$A=A_0e^{rt}\\3000=6000e^{r(3)}$

$0.5=e^{(24.5)r}$

After substituting, we divided by 6000 to each side to get 0.5 on the left. Now to solve for r, we will take the natural log of both sides and use log rules to isolate r.

$ln 0.5=ln e^{(3)r}\\ln 0.5=3r (ln e)\\\frac{ln0.5}{3} =r$