I need help with Algebra 2, please help!!!

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0

4 years ago

A basket of food is sufficient to feed 10 persons for 3 days. How many days would it last for 5 persons?

Norma-Jean [14]

Answer:

6 days the food would last for 5 persons.

Step-by-step explanation:

It would take double the days normally, as less people.

10 persons → 3 days

1 persons → (3 * 10) days

→ 30 days

5 persons → 30/5

→ 6 days

4 0

3 years ago

Read 2 more answers

Passes through (2,-5), parallel to with (1,3) and (4,5)

JulijaS [17]

so first you need to find the slop between (1,3) and (4,5)

5-3/ 4-1 = 2/ 3= m

using that slope you put it into the slope intercept formula

y=mx+b

use the coordinate (2,-5)

-5= 2/3 (2)+b

-5= 1.33+b subtract 1.33 on both sides

-6.33=b

plug b and m into the slope intercept formula

ANSWER: y=2/3x-6.33

6 0

3 years ago

Help meh ples and thank you. :p

Sladkaya [172]

Answer:

38.465

Step-by-step explanation:

Area of circle =

$\pi \: r {}^{2}$

diameter= 7

radius is half of the diameter

Radius =7÷2 = 3.5

Area = 3.14 × 3.5 × 3.5

= 38.465 inch squared

7 0

3 years ago

Find an equation for the line that passes through the points (-1,2) and (5,5)

sesenic [268]

Answer:

-The equation for the problem:

$y = \frac{1}{2}x + \frac{5}{2}$

Step-by-step explanation:

-First, you need determine the slope of a line:

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ (where $(x_{1},y_{1})$ is the first coordinate and $(x_{2},y_{2})$ is the second coordinate).

-Use both points $(-1,2)$ and $(5,5)$ for the formula:

$m = \frac{5-2}{5+1}$

$m = \frac{3}{6}$

$m = \frac{1}{2}$

-After you have found the slope, use the point-slope formula and use the slope $\frac{1}{2}$ and the first coordinate, which is $(-1,2)$ , to solve the equation and put it in slope-intercept form:

$y-y_{1} = m (x-x_{1})$

$y-2 = \frac{1}{2} (x+1)$

-Solve:

$y-2 = \frac{1}{2} (x+1)$

$y-2 = \frac{1}{2}x + \frac{1}{2}$

$y = \frac{1}{2}x + \frac{5}{2}$

So, therefore the equation is $y = \frac{1}{2}x + \frac{5}{2}$ .

8 0

4 years ago