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DIA [1.3K]
3 years ago
7

What is the equation of the quadratic graph with a focus of (3, -1) and a directrix of y = 1?

Mathematics
1 answer:
nydimaria [60]3 years ago
7 0

Answer:

The equation of the quadratic graph is y =  -\frac{1}{4} (x - 3)²

Step-by-step explanation:

The standard form of the equation of the quadratic graph is

(x - h)² = 4p(y - k), where

  • The vertex of the parabola is (h, k)
  • The focus is (h, k + p)
  • The directrix is at y = k - p  

∵ The focus is (3, -1)

∵ The focus is (h, k + p)

→ Compare them

∴ h = 3

∴ k + p = -1 ⇒ (1)

∵ The directrix is at y = 1

∵ The directrix is at y = k - p

→ Compare them

∴ k - p = 1 ⇒ (2)

→ Add equations (1) and (2) to eliminate p

∵ (k + k) + (p - p) = (-1 + 1)

∴ 2k + 0 = 0

∴ 2k = 0

→ Divide both sides by 2

∴ K = 0

→ Substitute the value of k in equation (1) to find p

∵ 0 + p = -1

∴ p = -1

→ Substitute the values of h, k, and p in the form of the equation above

∵ (x - 3)² = 4(-1)(y - 0)

∴ (x - 3)² = -4(y)

∴ (x - 3)² = -4y

→ Divide both sides by -4

∴ -\frac{1}{4} (x - 3)² = y

→ Switch the two sides

∴ y =  -\frac{1}{4} (x - 3)²

∴ The equation of the quadratic graph is y =  -\frac{1}{4} (x - 3)²

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I hope my answer has come to your help. God bless and have a nice day ahead!
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4 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
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Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

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P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

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1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

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We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

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Step-by-step explanation:

1. To solve this problem you must apply the formula for calculate the area of a circle, which is shown below:

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Where<em> r</em> is the radius of the circle.

2. The diameter of a circle is:

D=\frac{r}{2}

Where <em>r</em> is the radius of the circle.

3. Therefore, keeping this on mind, you have that the light-collecting area of a  50-meter keck telescope is:

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A_2=(\frac{10m}{2})^{2}\pi=78.53m^{2}

5. Divide A_1 by A_2, then:

\frac{1963.49m^{2}}{78.53m^{2}}=25

6. Therefore, its light-collecting area would be 25 times greater than that of the 10-meter keck telescope.

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