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2 years ago

Y=x^(2) -2x - 3 and y = x - 3

Mathematics

1 answer:

Oliga [24]2 years ago

5 0

Answer:

(0, -3) and (3, 0)

Explanation:

1st equation: y = x² -2x - 3

2nd equation : y = x - 3

Using substitution method:

x² -2x - 3 = x - 3 exchange sides

x² - 2x - x - 3 + 3 = 0 simplify

x² - 3x = 0 take common factor

x(x - 3) = 0 simplify

x = 0, 3

Solve for y:

when x is 0, y = x - 3 = 0 - 3 = -3

when x is 3, y = x - 3 = 3 - 3 = 0

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3 years ago

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If demand function is Qd=150+10p and supply function is Qs=300-2p, find the equilibrium price.

olga nikolaevna [1]

Answer:

$qd = qs$

150+10p=300-2p

10p+2p=300-150

Step-by-step explanation:

12p=150

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The equilibrium price is 12.5 or 13

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2 years ago

Pachacha [2.7K]

Answer:

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2 years ago

-10(10x-12)=-9(-9x-2)-5

Viefleur [7K]

10(10x−12)=−9(−9x−2)−5
Step 1: Simplify both sides of the equation.
−10(10x−12)=−9(−9x−2)−5
(−10)(10x)+(−10)(−12)=(−9)(−9x)+(−9)(−2)+−5(Distribute)
−100x+120=81x+18+−5
−100x+120=(81x)+(18+−5)(Combine Like Terms)
−100x+120=81x+13
−100x+120=81x+13
Step 2: Subtract 81x from both sides.
−100x+120−81x=81x+13−81x
−181x+120=13
Step 3: Subtract 120 from both sides.
−181x+120−120=13−120
−181x=−107
Step 4: Divide both sides by -181.
−181x
−181
=
−107
−181
x=
107
181
Answer:
x=
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3 years ago

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The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

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3 years ago