All categories

3 years ago

Write an equation in point slope form for each line Slope is 3 and (-2,-1) is on the line

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

6 0

Answer:

$\huge \blue{y + 1 = 3(x + 2) } \\$

Step-by-step explanation:

To find an equation of a line in point slope form when given the slope and a point we use the formula

$y - y_1 = m(x - x_1)$

From the question we have the final answer as

$y + 1 = 3(x + 2)$

Hope this helps you

You might be interested in

2(7x-34) = 22<br> Find X<br> Brainliest

Zepler [3.9K]

Answer: x = 45/7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

8. How do William James's theories on memory relate to modern ideas about short-term and

FromTheMoon [43]

Answer:

As it sounds, the idea itself.

Step-by-step explanation:

3 0

3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=9-1=8$

The Confidence interval is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and the critical values are:

$\chi^2_{\alpha/2}=20.09$

$\chi^2_{1- \alpha/2}=1.65$

And replacing into the formula for the interval we got:

$\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0

3 years ago

Round 1.965 to the nearest tenth. please

Alona [7]

I’m pretty sure it’s 2.0 have a nice day!

6 0

3 years ago

To figure this out, fill in the blanks and do the math:

Anit [1.1K]

1 is 60. 2 is 60. 3 is 24. 4 is 365.25. 5 Is 186282. 6 is 5879000000000 or 5.879 X 10^12.

7 0

3 years ago