The polynomial y=x^2-2x 1 has a repeated factor. true or false?

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

The tv costs $850 you pay 20% , how much is left to pay

kogti [31]

The answer would be 680 because 20% of 850 is 170 so 850-170=680.

3 0

2 years ago

Read 2 more answers

Rochelle works part-time at the college museum, which pays her at a rate of $10 per hour. She also makes small bracelets that sh

katrin [286]

We know for our problem that Rachel makes $10 per hour. Since $x$ represent the number of hours that she works, $10x$ will be the total amount that she will make for working $x$ hours. We also know that she sells bracelets for $5 each. Since $y$ represents the number of of bracelets that she sells, $5y$ will be her total revenue for selling $y$ bracelets. We also know that she needs to earn at least $200 a week to cover her expenses, so the sum of $10x$ and $5y$ must be equal or greater than 200:

$10x+5 \geq 200$

We can conclude that the graphs that shows the inequality that represents this situation, with its solution region shaded is: