All categories

3 years ago

Show that (f o g)(x) = (g o f)(x) = x

Mathematics

1 answer:

cluponka [151]3 years ago

6 0

Answer:

$- \frac{2[1 - x]}{3} = g[f(x)] \\ \\ \frac{3x}{2 - x} = f[g(x)]$

Step-by-step explanation:

They are not.

For the g[f(x)] function, you substitute ³/ₓ ₋ ₁ from the f(x) function in for x in the g(x) function to get this:

$\frac{2}{ \frac{3}{x - 1}}$

Then, you bring x - 1 to the top while changing the expression to its conjugate [same expressions with opposite symbols]:

$- \frac{2[1 - x]}{3}$

You could also do this [attaching another negative would make that positive].

For the f[g(x)] function, ²/ₓ from the g(x) function for x in the f(x) function to get this:

$\frac{3}{ \frac{2}{x} - 1}$

Now, if you look closely, ²/ₓ is written as 2x⁻¹, and according to the Negative Exponential Rule, you bring the denominator to the numerator while ALTERING THE INTEGER SYMBOL FROM NEGATIVE TO POSITIVE:

$\frac{3x}{2 - x}$

When this happens, x leaves the two and gets attached to the three, and 1 gets an x attached to it.

I am joyous to assist you anytime.

You might be interested in

Judy earns $65 for working five hours. find the constant rate of variation

d1i1m1o1n [39]

Answer:

which principle prevents a brach from abusing its power

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Prove this ~ cosx/1-tanx + sinx/1-cotx = cosx + sinx

bearhunter [10]

The solution is attached in the image below. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

3 0

3 years ago

he production of pipes has a mean diameter of 3.25 inches and a standard deviation of .15 inches. The shape of the distribution

o-na [289]

Answer:

The probability that a randomly chosen part has diameter of 3.5 inches or more is 0.9525

Step-by-step explanation:

$Mean = \mu = 3.25 inches$

Standard deviation = $\sigma = 0.15 inches$

We are supposed to determine the probability that a randomly chosen part has diameter of 3.5 inches or more

$P(Z \geq 3.5)=1-P(z$

Refer the z table for p value

$P(Z \geq 3.5)=1-0.0475\\P(Z \geq 3.5)=0.9525$

Hence the probability that a randomly chosen part has diameter of 3.5 inches or more is 0.9525

7 0

3 years ago

How many dimensions does a point have

vladimir2022 [97]

Point has no dimensions !!

6 0

3 years ago

Read 2 more answers

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago