Roster form {-2,-1,0,1,2,3} set builder form

3. Simplify. * 5m3 + 2m– 7m3 – 8m

Oksanka [162]

Step-by-step explanation:

5m³ + 2m - 7m³ - 8m

= -2m³ -6m

= -2m(m² + 3)

4 0

3 years ago

Can someone give me some tips of how to study. I have already tried writing everything in my lesson but it doesn't seem to work

Ainat [17]

Study all notes, reread the chapters again. Have someone ask questions on the chapters page by page. This always has worked for me. Plus try to do this again the night before the test. You will be surprised how much you can remember by doing it again the night before the test. Hope this helps.

8 0

3 years ago

On the basis of data collected during an experiment, a biologist found that the growth of a fruit fly population (Drosophila) wi

Ket [755]

Answer:

(a). 15

(b). 78

Step-by-step explanation:

Growth of the population of a fruit fly is modeled by

N(t) = $\frac{600}{1+39e^{-0.16t} }$

where t = number of days from the beginning of the experiment.

(a). For t = 0 [Initial population]

N(0) = $\frac{600}{1+39e^{-0.16\times 0} }$

= $\frac{600}{1+39}$

= $\frac{600}{40}$

= 15

Initial population of the fruit flies were 15.

(b).Population of the fruit fly colony on 11th day.

N(11) = $\frac{600}{1+39e^{-0.16\times 11} }$

= $\frac{600}{1+39e^{-1.76} }$

= $\frac{600}{1+39\times 0.172 }$

= $\frac{600}{1+6.71}$

= $\frac{600}{7.71}$

= 77.82

≈ 78

On 11th day number of fruit flies colony were 78.

3 0

4 years ago

Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt

-Dominant- [34]

Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is tested if it does not outlast by more than 2 hours, that is, the subtraction is not more than 2:

$H_0: \mu_B - mu_A \leq 2$

At the alternative hypothesis, it is tested if it outlasts by more than 2 hours, that is:

$H_1: \mu_B - \mu_A > 2$

The sample means are: $\mu_A = 8.65, \mu_B = 11.23$

The standard deviations for the samples are $s_A = s_B = 0.67$

Hence, the standard errors are:

$s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934$

The distribution of the difference has mean and standard deviation given by:

$\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58$

$s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 2$ is the value tested at the hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{2.58 - 2}{0.2735}$

$t = 2.12$

The critical value for a right-tailed test, as we are testing if the subtraction is greater than a value, with a 0.05 significance level and 12 + 12 - 2 = 22 df is given by $t^{\ast} = 1.71$

Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0

3 years ago

Select the correct answer. Which graph shows a function and its inverse?

labwork [276]

Answer:

See below

Step-by-step explanation:

If a function is bijective and 1-to-1, then it will have an inverse function. Consequentially, they will be symmetrical about the line $y=x$ , which is a diagonal line passing through the origin at a 45 degree angle.

None of the graphs look correct though, but it also seems that some options are cut out, so make sure to choose the correct graph given the characteristics I've previously described.

6 0

2 years ago

Read 2 more answers