Which other figure is a part of line CD?

In a group of 60 students, 18 students take Algebra I, 20 students take Algebra II, and 10 students take both subjects. How many

andreyandreev [35.5K]

Out of the total 60 students 18 take Algebra 1 and 20 take Algebra II. 10 take both. So out of the 18 taking algebra 10 are taking Algebra II. If we don't count that 10 because they are already counted in Algebra II count. So 28 students are taking either course. Therefore 60-28 = 32 students who are not taking either of these courses

3 0

3 years ago

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Look at the inequality below:

malfutka [58]

I believe that answer is C

4 0

3 years ago

Central= 127 Radius= 7. What is the length of the indicated arc (x)

tia_tia [17]

Given:

$\begin{gathered} \theta\text{ = 127} \\ r\text{ = 7 mm} \end{gathered}$

Required:

Length of an arc x.

Explanation:

Length of an arc is given as,

$Length\text{ of an arc =}\frac{\theta}{360}\times2\pi\text{r}$

Length of an arc is calculated as,

$\begin{gathered} Length\text{ of an arc = }\frac{127}{360}\times2\times3.14\times7 \\ Length\text{ of an arc = 0.3528 }\times2\times3.14\times7 \\ Length\text{ of an arc = 15.5091 mm} \\ \end{gathered}$

Answer:

Thus the value of the length of an arc is 15.5091 mm

5 0

2 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

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

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

7 0

3 years ago

What do you mean by the term standard unit ?

IRINA_888 [86]

A standard unit of measurement is a quantifiable language that helps everyone understand the association of the object with the measurement. It is expressed in inches, feet, and pounds, in the United States, and centimeters, meters, and kilograms in the metric system.

3 0

4 years ago

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