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Let f(x)=x^2 +2 and g(x)=1-3x. find each function value: (fg)(-1)

sammy [17]

Answer:

12

Step-by-step explanation:

so basically

(x^2+2)*(1-3x)=-3x^3+x^2-6x+2

now plug in -1

-3(-1)^3+(-1)^2-6(-1)+2=12

hope this helped :)

6 0

3 years ago

Read 2 more answers

Find an equation perpendicular to x=1 and passing through (8,-9)

valina [46]

Answer:

$y=-9$

Step-by-step explanation:

We want to find an equation of a line that's perpendicular to x=1 that also passes through the point (8,-9).

Note that x=1 is a <em>vertical line </em>since x is 1 no matter what y is.

This means that if our new line is perpendicular to the old, then it must be a <em>horizontal line</em>.

So, since we have a horizontal line, then our equation must be our y-value of our point.

Our y-coordinate of our point (8,-9) is -9.

Therefore, our equation is:

$y=-9$

And this is in standard form.

And we're done!

3 0

3 years ago

1. Answer the following questions. (a) Check whether or not each of f1(x), f2(x) is a legitimate probability density function f1

kobusy [5.1K]

Answer:

f1 ( x ) valid pdf . f2 ( x ) is invalid pdf

k = 1 / 18 , i ) 0.6133 , ii ) 0.84792

Step-by-step explanation:

Solution:-

A) The two pdfs ( f1 ( x ) and f2 ( x ) ) are given as follows:

$f_1(x) = \left \{ {{0.5(3x-x^3) } .. 0 < x < 2 \atop {0} } \right. \\\\f_2(x) = \left \{ {{0.3(3x-x^2) } .. 0 < x < 2 \atop {0} } \right. \\$

- To check the legitimacy of a continuous probability density function the area under the curve over the domain must be equal to 1. In other words the following:

$\int\limits^a_b {f_1( x )} \, dx = 1\\\\ \int\limits^a_b {f_2( x )} \, dx = 1\\$

- We will perform integration of each given pdf as follows:

$\int\limits^a_b {f_1(x)} \, dx = \int\limits^2_0 {0.5(3x - x^3 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - 0.125x^4 ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - 0.125*(16) ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 2 ] = 1\\$

$\int\limits^a_b {f_2(x)} \, dx = \int\limits^2_0 {0.5(3x - x^2 )} \, dx \\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75x^2 - \frac{x^3}{6} ]\limits^2_0\\\\\int\limits^a_b {f_1(x)} \, dx = [ 0.75*(4) - \frac{(8)}{6} ]\\\\\int\limits^a_b {f_1(x)} \, dx = [ 3 - 1.3333 ] = 1.67 \neq 1 \\$

Answer: f1 ( x ) is a valid pdf; however, f2 ( x ) is not a valid pdf.

B)

- A random variable ( X ) denotes the resistance of a randomly chosen resistor, and the pdf is given as follows:

$f ( x ) = kx$ if 8 ≤ x ≤ 10

0 otherwise.

- To determine the value of ( k ) we will impose the condition of validity of a probability function as follows:

$\int\limits^a_b {f(x)} \, dx = 1\\$

- Evaluate the integral as follows:

$\int\limits^1_8 {kx} \, dx = 1\\\\\frac{kx^2}{2} ]\limits^1^0_8 = 1\\\\k* [ 10^2 - 8^2 ] = 2\\\\k = \frac{2}{36} = \frac{1}{18}$ ... Answer

- To determine the CDF of the given probability distribution we will integrate the pdf from the initial point ( 8 ) to a respective value ( x ) as follows:

$cdf = F ( x ) = \int\limits^x_8 {f(x)} \, dx\\\\F ( x ) = \int\limits^x_8 {\frac{x}{18} } \, dx\\\\ F ( x ) = [ \frac{x^2}{36} ] \limits^x_8\\\\F ( x ) = \frac{x^2 - 64}{36}$