3x^3-24x^2 factor completley

Match each graph with the correct slope of the line. drag and drop the answer

OLEGan [10]

Answer/Step-by-step explanation:

✔️Slope of the first graph:

Using two points on the line, (0, 1) and (3, 2),

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{3 - 0} = \frac{1}{3}$

Slope = ⅓

✔️Slope of the second graph:

Using two points on the line, (0, 0) and (1, 1),

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 0}{1 - 0} = \frac{1}{1} = 1$

Slope = 1

✔️Slope of the third graph:

Using two points on the line, (0, 1) and (2, 2),

$slope (m) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{2 - 0} = \frac{1}{2}$

Slope = ½

3 0

3 years ago

Read 2 more answers

find the equation of a line that passes through (-3,1) and (-1,-2) and write it in standard form. please help me !

Savatey [412]

Find the slope first
slope=(2-1)/(-1+3)=1/4=.25

now use the slope and either point to write in point slope form
y-1=.25(x-2)
y-1=¼x-½

to rationalize all the denominators multiply by the last common multiple, here we get 4
4y-4=x-2

then put in the correct order

x-4y=-2

8 0

3 years ago

2X+2Y=48<br> 3X+Y=40 <br><br> What is X and Y

marshall27 [118]

Answer:

x=8

y=16

Step-by-step explanation:

2x+2y=48

3x+y=40

Ther is many ways to solve this, one of them is clearance:

WE CLEAR AN UNLOCKED (X) AND REPLACE IT IN THE OTHER

x= (40-y)/3

2[(40-y)/3] +2y=48

(80-2y)/3 +2y=48

(80-2y+6y)/3=48

80+4y=48(3)

4y=144

y= 144/4

y=16

x=(40-y)/3

x=(40-16)/3

x=24/3

x=8

4 0

3 years ago

THIS IS 100 POINTS SOMEBODY PLEASEEEEEEEEEEEEEE HELP MEEEEE

Nookie1986 [14]

Answer: Option #4: the 0.2 could be changed to 2.

Explanation:

The table is designed to make the concept of percentages more understandable. The table first allocates the total amount of 70 in terms of its "fifths," or 20% parts. Then it expresses these parts in the last row of the table, showing that 20% of 70 is 14 and all the five parts sum up to 70 again.

Now, Mikel is supposed to express 40% of the amount. So she writes (incorrectly): 0.2 * 14. This statement needs to be changed to 2 * 14 (two times 14 = 28), to correspond to 2 times 20%, or 40% of 70.

This is reflected in the last (fourth) option "The 0.2 in the expression could be changed to 2."

Option #3 is incorrect because changing 14 to 70, will result in an incorrect number (2.8).

Options 1 and 2 are similarly incorrect (as can be easily verified)

3 0

3 years ago

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Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test

slamgirl [31]

Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

The 95% confidence interval would be given by (96.625;100.915)

a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

c) The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

$\bar X=98.77$ represent the sample mean

$s=4.77$ represent the sample standard deviation

n=19 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$

$98.77+1.96\frac{4.77}{\sqrt{19}}=100.915$

So on this case the 95% confidence interval would be given by (96.625;100.915)

Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{18}$

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part c

If the system of hypothesis on this case are:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu > 100$

The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

7 0

3 years ago