Can you put 4/0 on a number line

Simplify 15 + 6(x + 1).

expeople1 [14]

The answer is: 6x+21

5 0

3 years ago

Read 2 more answers

Fine the probability of not rolling factor of 4 on both dice

Marianna [84]

Answer:

the answer is 6

Step-by-step explanation:

5 0

3 years ago

What would be a equation for (5,2) and (0,7)?

Genrish500 [490]

$\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{7}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{5}}}\implies \cfrac{5}{-5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-1}(x-\stackrel{x_1}{5}) \\\\\\ y-2=-x+5\implies y = -x+7$

6 0

3 years ago

Trukkkkkkkkkk that hurt like a butt cheek on a stick

kiruha [24]

✨Vines vines vines vines vines vines vines✨

3 0

2 years ago

An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume

Lynna [10]

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

Part a

$\bar X=21$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=52$ represent the population standard deviation

n=350 represent the sample size

99% confidence interval

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

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

Part b

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{350}}=7.17$

Part c

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{100}}=13.42$

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

6 0

3 years ago