1. Who conducted the study described in the article?

Which of the following expressions is equivalent to the expression 5(4n - 11) - 3n + 20?

WINSTONCH [101]

17n-35 would be the correct answer

7 0

2 years ago

4. Tina wants to secure sequins on a piece of felt shaped like a rectangle with a length of 10 cm and width of 21 cm. What is th

mote1985 [20]

First we find the area 10 x 21 = 210 then we multiply 210 x .70 and we get 147.00 total cost. d. 147.00

4 0

2 years ago

Read 2 more answers

HELP PLEASE. A pizza shop charges customers based on the area of the pizza they order. The cost of a pizza in dollars can be rep

san4es73 [151]

For this case we have that the cost of each pizza is given by the following function:

$S (r) = 0.1 \pi * r ^ 2$

Where:

r: It's the radius of the pizza

IF a pizza cost $26.32 we have:

$26.32 = 0.1 \pi * r ^ 2$

We cleared the radius, using $\pi = 3.14$ :

$\frac {26.32} {0.1} = \pi * r ^ 2\\263.2 = 3.14r ^ 2\\\frac {263.2} {3.14} = r ^ 2\\83.82 = r ^ 2\\r = \pm \sqrt {83.82}$

We choose the positive value:

$r = 9.15$

So, the radius of the pizza is $9.16 \ in$

Then, the diameter is: $2r = 2 * 9.15 = 18.30 \ in$

Finally, the diameter of the pizza is: $18.30 \ in$

Answer:

Option B

4 0

2 years ago

Read 2 more answers

6.In a sample of 131 women with cosmetic dermatitis from using eye shadow, 12 were diagnosed with a nickel allergy. In a sample

aivan3 [116]

Answer:

a) $0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

b) $\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

c) For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

Step-by-step explanation:

Part a

The estimated proportion of women with cosmetic dermatitis from using eye shadow is given by:

$\hat p =\frac{12}{131}= 0.0916$

The confidence interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing we got:

$0.0916 - 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.0422$

$0.0916 + 1.96\sqrt{\frac{0.0916(1-0.0916)}{131}}=0.1410$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using eye shadow at 95% of confidence is between (0.0422 and 0.1410)

Part b: A 95% confidence interval for the women with cosmetic dermatitis from using mascara

$\hat p = \frac{25}{250}= 0.1$

$0.1 - 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.0628$

$0.1 + 1.96\sqrt{\frac{0.1(1-0.1)}{250}}=0.1372$

So then we can conclude that the true proportion of women with cosmetic dermatitis from using mascara at 95% of confidence is between (0.0628 and 0.1372)

Part c: Suppose you are informed that the true proportion with a nickel allergy for one of the two groups (eye shadow or mascara) is .12. Can you determine which group is referenced? Explain.

For this case we see that both confidence intervals contains the value of 0.12 so then we can't conclude that only one group is referenced at the significance level of 0.05 used.

7 0

2 years ago

Employees in the marketing department of a large regional restaurant chain are researching the amount of money that households i

kap26 [50]

Answer:

The z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

Step-by-step explanation:

The data provided is as follows:

25 to 34 45 to 54

1329 2268

1906 1965

2426 1149

1826 1591

1239 1682

1514 1851

1937 1367

1454 2158

Compute the mean and standard deviation for the group "25 to 34" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [1329+1906+...+1454]=\frac{13631}{8}=1703.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1086710.875}=394.01$

Compute the z-score for the group "25 to 34" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1703.875}{394.01}=0.3734\approx 0.37$

Compute the mean and standard deviation for the group "45 to 54" as follows:

$\bar x=\frac{1}{n}\sum x=\frac{1}{8}\times [2268+1965+...+2158]=\frac{14031}{8}=1753.875\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 1028888.875}=383.39$

Compute the z-score for the group "45 to 54" as follows:

$z=\frac{x-\bar x}{s}=\frac{1851-1753.875}{383.39}=0.25333\approx 0.25$

Thus, the z-score for the group "25 to 34" is 0.37 and the z-score for the group "45 to 54" is 0.25.

8 0

2 years ago