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3 years ago

Find the inverse of the given function.

Mathematics

1 answer:

Bogdan [553]3 years ago

7 0

Answer:

The inverse is

Step-by-step explanation:

$y=3\sqrt{2x^{2} -5} + 3$

To find the inverse of the function interchange the terms that's x becomes y and y becomes x

We have

$x=3\sqrt{2y^{2}-5 }+3$

Now solve for y

Move 3 to the other side of the equation

$3\sqrt{2y^{2}-5 } = x-3$

Divide both sides by 3

We have

$\sqrt{2y^{2}-5 } =\frac{x-3}{3}$

square both sides of the equation to remove the square root

That's

$2y^{2}-5 = (\frac{x-3}{3}) ^{2}$

Move 5 to the other side of the equation

$2y^{2} = (\frac{x-3}{3}) ^{2}+5$

Divide both sides by 2

We have

$y^{2} = \frac{(\frac{x-3}{3}) ^{2}+5}{2}$

Find the square root of both sides

We have the final answer as

$y =\sqrt{\frac{(\frac{x-3}{3}) ^{2}+5}{2}}$

Hope this helps you

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F(x) = x + 2 g(x) = x - 4 (f9)(x) =

san4es73 [151]

Answer:

(fºg)(x) = x -2

Step-by-step explanation:

f(x) = x + 2

g(x) = x - 4

(fºg)(x) = ( x - 4) + 2

(fºg)(x) = x -2

3 0

2 years ago

Assume that the random variable X is normally distributed with mean u = 45 and standard deviation o = 14.

larisa86 [58]

Answer:

P(57 < X < 69) = 0.1513

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 45, \sigma = 14$

Find P(57 < X < 69):

This is the pvalue of Z when X = 69 subtracted by the pvalue of Z when X = 57. So

X = 69

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{69 - 45}{14}$

$Z = 1.71$

$Z = 1.71$ has a pvalue of 0.9564

X = 57

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{57 - 45}{14}$

$Z = 0.86$

$Z = 0.86$ has a pvalue of 0.8051

0.9564 - 0.8051 = 0.1513

P(57 < X < 69) = 0.1513

4 0

3 years ago

Find the length of AC again

Bingel [31]

Answer:

Step-by-step explanation:

side Ac is the opposite of angle B, therefore use the sin ratio to find ac since the hypotenuse is also given

sin B=opposite/hypotenuse

sin 27=ac/15

ac=sin27×15

=6.81

I hope this helps

5 0

2 years ago

Read 2 more answers

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.

- 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:

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