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miskamm [114]
3 years ago
13

Suppose a parabola has vertex (6,5) and also passes through the point (7,7). Write the equation of the parabola in vertex form.

Mathematics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

Choice B: y = 2\, (x - 6)^{2} + 5.

Step-by-step explanation:

For a parabola with vertex (h,\, k), the vertex form equation of that parabola in would be:

\text{$y = a\, (x - h)^{2} + k$ for some constant $a$}.

In this question, the vertex is (6,\, 5), such that h = 6 and k = 5. There would exist a constant a such that the equation of this parabola would be:

y = a\, (x - 6)^{2} + 5.

The next step is to find the value of the constant a.

Given that this parabola includes the point (7,\, 7), x = 7 and y = 7 would need to satisfy the equation of this parabola, y = a\, (x - 6)^{2} + 5.

Substitute these two values into the equation for this parabola:

7 = (7 - 6)^{2}\, a + 5.

Solve this equation for a:

7 = a + 5.

a = 2.

Hence, the equation of this parabola would be:

y = 2\, (x - 6)^{2} + 5.

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The solutions to the questions are

  • The probability that X is between 2 and 4 is 0.314
  • The probability that X exceeds 3 is 0.199
  • The expected value of X is 2
  • The variance of X is 2

<h3>Find the probability that X is between 2 and 4</h3>

The probability density function is given as:

f(x)= xe^ -x for x>0

The probability is represented as:

P(x) = \int\limits^a_b {f(x) \, dx

So, we have:

P(2 < x < 4) = \int\limits^4_2 {xe^{-x} \, dx

Using an integral calculator, we have:

P(2 < x < 4) =-(x + 1)e^{-x} |\limits^4_2

Expand the expression

P(2 < x < 4) =-(4 + 1)e^{-4} +(2 + 1)e^{-2}

Evaluate the expressions

P(2 < x < 4) =-0.092 +0.406

Evaluate the sum

P(2 < x < 4) = 0.314

Hence, the probability that X is between 2 and 4 is 0.314

<h3>Find the probability that the value of X exceeds 3</h3>

This is represented as:

P(x > 3) = \int\limits^{\infty}_3 {xe^{-x} \, dx

Using an integral calculator, we have:

P(x > 3) =-(x + 1)e^{-x} |\limits^{\infty}_3

Expand the expression

P(x > 3) =-(\infty + 1)e^{-\infty}+(3+ 1)e^{-3}

Evaluate the expressions

P(x > 3) =0 + 0.199

Evaluate the sum

P(x > 3) = 0.199

Hence, the probability that X exceeds 3 is 0.199

<h3>Find the expected value of X</h3>

This is calculated as:

E(x) = \int\limits^a_b {x * f(x) \, dx

So, we have:

E(x) = \int\limits^{\infty}_0 {x * xe^{-x} \, dx

This gives

E(x) = \int\limits^{\infty}_0 {x^2e^{-x} \, dx

Using an integral calculator, we have:

E(x) = -(x^2+2x+2)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x) = -(\infty^2+2(\infty)+2)e^{-\infty} +(0^2+2(0)+2)e^{0}

Evaluate the expressions

E(x) = 0 + 2

Evaluate

E(x) = 2

Hence, the expected value of X is 2

<h3>Find the Variance of X</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Where:

E(x^2) = \int\limits^{\infty}_0 {x^2 * xe^{-x} \, dx

This gives

E(x^2) = \int\limits^{\infty}_0 {x^3e^{-x} \, dx

Using an integral calculator, we have:

E(x^2) = -(x^3+3x^2 +6x+6)e^{-x}|\limits^{\infty}_0

Expand the expression

E(x^2) = -((\infty)^3+3(\infty)^2 +6(\infty)+6)e^{-\infty} +((0)^3+3(0)^2 +6(0)+6)e^{0}

Evaluate the expressions

E(x^2) = -0 + 6

This gives

E(x^2) = 6

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = 6 - 2^2

Evaluate

V(x) = 2

Hence, the variance of X is 2

Read more about probability density function at:

brainly.com/question/15318348

#SPJ1

<u>Complete question</u>

A random variable X with a probability density function f(x)= xe^ -x for x>0\\ 0& else

a. Find the probability that X is between 2 and 4

b. Find the probability that the value of X exceeds 3

c. Find the expected value of X

d. Find the Variance of X

7 0
2 years ago
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