Ask question

Ask question

All categories

Vedmedyk [2.9K]

4 years ago

6

The directions on a sewing pattern say to cut an extra 15% of fabric to account for error. Chloe needs 0.75 yard of fabric to ma

ke a skirt, and she cuts 0.1125 yard. Did Chloe cut the correct amount of fabric? Why or why not?

1 answer:

ivann1987 [24]4 years ago

4 0

She was correct to multiply .75 by .15 to get .1125, but she forgot to add that value to .75 to get the total amount of fabric needed.

You might be interested in

2/3-1/4 I want the answer please

klemol [59]

You want to make the fractions have the same denominator.
2/3 = 8/12
1/4 = 3/12
now subtract the numerator and keep the denominator the same...
8/12 - 3/12 = 5/12
5 1/2 is the answer

6 0

3 years ago

Read 2 more answers

Rewrite 14/15 and 9/10

erik [133]

I think I know what your asking so the answer would be 27/30 and 28/30

6 0

3 years ago

Read 2 more answers

Please helo i will make brainless for correct answers!

elena-s [515]

False is the correct answer

7 0

4 years ago

Read 2 more answers

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

For every how many degrees will a pentagon have rotational symmetry?

igor_vitrenko [27]

A. 72 degrees (i hope this helped)

7 0

3 years ago

Read 2 more answers