What is the domain and range of y=6

Which statement is true of the function f(x) = -3/x? Select three options.

melamori03 [73]

<h2>Answer:</h2>

We have the following function:

$f(x)=-\frac{3}{x}$

The graph of this function has been plotted below. So lets analyze each statement:

<h3>1. The function is always increasing. </h3><h3>False</h3>

As you can see x increases from -∞ to 0 and decreases from 0 to +∞

<h3>2. The function has a domain of all real numbers. </h3><h3>False</h3>

The function is undefined for $x=0$ since x is in the denominator.

<h3>3. The function has a range of {yl- </h3>

Statement is unclear but the range is the set of all real numbers except zero.

<h3 /><h3>4. The function is a reflection of y = 3. </h3><h3>False</h3>

The function is a reflection in the x axis of the function $g(x)=\frac{3}{x}$

<h3>5. The function passes through the point (3,-27).</h3><h3>False</h3>

This is false since:

$f(3)=-1\neq -27$

Note. As you can see those statements are false, so any of them is true, except item 3 that is unclear.

7 0

3 years ago

Read 2 more answers

A single-celled organism is represented below. Structure X carries out a function most similar to which structure in a human?

tester [92]

Answer: lung

Step-by-step explanation:

Attachedfile/picture shows the structure.

Respiration is a process of degradation of complex organic compound with the production of carbon dioxide, water and energy.

Respiration involves two phases, which are;

(1). External Respiration or Breathing: this is a process in which

animals take oxygen in and release carbon dioxide.

(2). Internal Respiration or Cellular Respiration: this process involve the release of energy from food substance with the release of carbondioxide and and water.

Single celled animals or unicellular animals such as amoeba exchange gases through cell surface. The STRUCTURE X IS THE PLASMA MEMBRANE. There is absorption of of Oxygen from the surrounding air or water,hence, giving out carbondioxide through plasma membrane by Diffusion.

PS: Lung is used in respiration process in Human

4 0

4 years ago

The length of the rectangle below is (2x-3) and the width is (x+7). Find the area of the rectangle in terms of x.

krek1111 [17]

Answer:

rgfb

Step-by-step explanation:

2 because if you divide

5 0

3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

A patient weighs 50 lbs on admission, and 48 lbs on day 3 of his hospital stay. Write a fraction, reduced to lowest terms, to ex

rusak2 [61]

Answer:

The fraction that express the weight patient has lost is $\frac{1}{25}$ .

Step-by-step explanation:

Given:

Weight of patient on admission= 50 lbs

Weight of patient on day 3 of his hospital stay = 48 lbs

We need to write the fraction reduced to lowest terms, to express the weight he has lost.

Solution:

First we will find the total weight lost.

total weight lost can be calculated by subtracting Weight of patient on day 3 of his hospital stay from Weight of patient on admission.

framing in equation form we get;

total weight lost = $50-48 = 2\ lbs$

Now we will find the fraction of weight he has lost.

To find the fraction of weight he has lost we will divided total weight lost from Weight of patient on admission.

framing in equation form we get;

fraction of weight he has lost = $\frac{2}{50}$

Reducing it in lowest term we get;

fraction of weight he has lost = $\frac{2}{2\times 25}= \frac{1}{25}$

Hence the fraction that express the weight patient has lost is $\frac{1}{25}$ .

5 0

3 years ago