Please help draw graph and do the table of x and y values

A class of students consists of 9 men and 51 women. Write a proper fraction to represent the part of the class that is women. Re

balandron [24]

Answer:

85%

Step-by-step explanation:

9+51 = 60

(51/3) ÷ (60/3) = 17/20

(17/20)% = 85%

5 0

2 years ago

For every 5 pages in a book Fred reads, Lonnie reads 4 pages. In 1 hour, they read a combined total of 36 pages. How many pages

nydimaria [60]

Answer:

Hence the number of pages read by Lonnie is 16 pages when they read a combined of 36 pages.

Step-by-step explanation:

The table that can be formed by the rate at which they study the number of pages of a book is:

Fred Lonnie total

5 4 9

10 8 18

15 12 27

20 16 36

That means when they read a combined of 36 pages.

Then the number of pages read by Fred is 20 pages.

and the number of pages read by Lonnie is 16 pages.

(

Also it could be done as :

Let Fred reads 'x' pages.

so, the number of pages read by Lonnie is: (4/5)x

Hence,

x+(4/5)x=36

on solving this linear equation we have:

x=20

and (4/5)x=16

)

Hence the number of pages read by Lonnie is 16 pages when they read a combined of 36 pages.

3 0

3 years ago

a pool is leaking water at a constant rate. The amount of water in the pool changes by -1 1/4 every hour. what is the total chan

malfutka [58]

The total change in the water in the pool after 2 1/2 hours given the rate of leakage is -3 1/8,

<h3>What are fractions?</h3>

A fraction is a non-integer. It is made up of numerators and denominators. A mixed fraction is made up of a whole number and a proper fraction. An example is 2 1/2.

<h3>What is the total change in the water?</h3>

The total change in the water = -1 1/4 x 2 1/2

= -5/4 x 5/2 = -25/8 = -3 1/8

To learn more about multiplication of fractions, please check: brainly.com/question/1114498

6 0

2 years ago

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA ba

PSYCHO15rus [73]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The attached Excel file 2013 NCAA BB Tournament shows the salaries paid to the coaches of 62 of the 68 teams in the 2013 NCAA basketball tournament (not all private schools report their coach's salaries). Consider these 62 salaries to be a sample from the population of salaries of all 346 NCAA Division I basketball coaches.

Question 1. Use the 62 salaries from the TOTAL PAY column to construct a 95% confidence interval for the mean salary of all basketball coaches in NCAA Division I.

xbar = $1,465,752

SD = $1,346,046.2

lower bound of confidence interval ________

upper bound of confidence interval _______

Question 2. Coach Mike Krzyzewski's high salary is an outlier and could be significantly affecting the confidence interval results. Remove Coach Krzyzewski's salary from the data and recalculate the 95% confidence interval using the remaining 61 salaries.

xbar = $1,371,191

SD = $1,130,666.5

lower bound of confidence interval _________

upper bound of confidence interval. ________

Answer:

Question 1:

lower bound of confidence interval = $1,124,027

upper bound of confidence interval = $1,807,477

Question 2:

lower bound of confidence interval = $1,081,512

upper bound of confidence interval = $1,660,870

Step-by-step explanation:

Question 1:

The sample mean salary of 62 couches is

$\bar{x} = 1,465,752$

The standard deviation of mean salary is

$s = 1,346,046.2$

The confidence interval for the mean salary of all basketball coaches is given by

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } )$

Where $\bar{x}$ is the sample mean, n is the sample size, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 62 - 1 = 61

From the t-table at α = 0.025 and DoF = 61

t-score = 1.999

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (\frac{1,346,046.2}{\sqrt{62} } ) \\\\CI = 1,465,752 \pm 1.999 \cdot (170948.04 ) \\\\CI = 1,465,752 \pm 341,725 \\\\LCI = 1,465,752 - 341,725 = 1,124,027 \\\\UCI = 1,465,752 + 341,725 = 1,807,477\\\\$

Question 2:

After removing the Coach Krzyzewski's salary from the data

The sample mean salary of 61 couches is

$\bar{x} = 1,371,191$

The standard deviation of the mean salary is

$s = 1,130,666.5$

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 61 - 1 = 60

From the t-table at α = 0.025 and DoF = 60

t-score = 2.001

So the required 95% confidence interval is

$CI = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (\frac{1,130,666.5}{\sqrt{61} } ) \\\\CI = 1,371,191 \pm 2.001 \cdot (144767 ) \\\\CI = 1,371,191 \pm 289,678.8 \\\\LCI = 1,371,191 - 289,678.8 = 1,081,512 \\\\UCI = 1,371,191 + 289,678.8 = 1,660,870\\\\$

6 0

3 years ago

Hello there!

djyliett [7]

2. f(x) = x - 2x² - 5 + 10x

=-2x² + 8x - 5

f'(x)= -4x + 8

4. y = 100(45x - 30 - 3x³ + 2x²)

= 100(-3x³ + 2x² + 45x -30)

= -300x³ + 200x² + 4500x - 3000

y' = -900x² + 400x + 4500

5 0

3 years ago