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3 years ago

14

Which linear inequality is represented by the graph?

2 answers:

Sholpan [36]3 years ago

5 0

The third one y is less than or equal too

Goryan [66]3 years ago

3 0

Answer:

y ≤ $\frac{2 }{ 3}$ + $\frac{1}{5}$ .

Step-by-step explanation:

Given : Graph .

To find : Which linear inequality is represented by the graph.

Solution : We have given graph with shaded down and solid line

It has solid line it mean inequality has equal to sign.

It is shaded down it mean inequality y has less than or equal to sign.

Slope = $\frac{y_{2}-y_{1} }{x_{2}- x_{1}}$ .

Slope = $\frac{2.2- 0.2 }{ 3- 0}$ .

Slope = $\frac{2 }{ 3}$ .

Y intercept = 0.2 = $\frac{2}{10}$ .

y intercept = $\frac{1}{5}$ .

Equation of line y = mx + b

Where m = slope , b = y intercept .

Plug the values of slope and y intercept .

y ≤ $\frac{2 }{ 3}$ + $\frac{1}{5}$ .

Therefore, y ≤ $\frac{2 }{ 3}$ + $\frac{1}{5}$ .

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Answer:

A. E ( U ) = 21.5454 , E ( F ) = 8.39333

B. M ( U ) = 17.0 , M ( F ) = 18.0

C. E ( U' ) = 17.0 , E ( F' ) = 7.95384

D. T ( U ) = 9.091% , T ( F ) = 6.667%

Step-by-step explanation:

Solution:-

- Two sample sets ( U ) and ( F ) that define the concentration ( EU/mg ) of endotoxin found in urban and farm homes as follows:

U: 6.0 5.0 11.0 33.0 4.0 5.0 80.0 18.0 35.0 17.0 23.0

F: 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 22.0 9.6 2.0 2.0 0.5

- To determine the mean of a sample E ( U ) or E ( F ) the following formula from descriptive statistics is used:

$E ( X ) = Sum ( X_i ) / n$

Where,

Xi : Data iteration

n: Sample size

Therefore,

$E ( U ) = \frac{Sum (U_i )}{n_u} \\\\E ( U ) = \frac{6.0 + 5.0 + 11.0 + 33.0 + 4.0+ 5.0 +80.0+ 18.0+ 35.0+ 17.0+ 23.0 }{11} \\\\E ( U ) = 21.54545\\\\E ( F ) = \frac{Sum (F_i )}{n_f} \\\\E ( F ) = \frac{2.0 + 15.0 + 12.0 + 8.0 + 8.0 + 7.0 + 6.0 + 19.0+ 3.0+ 9.8+ 22.0+ 9.6+ 2.0+ 2.0+ 0.5 }{15} \\\\E ( F ) = 8.39333$

- To determine the sample median we need to arrange the data for both samples ( U ) and ( F ) in ascending order as follows:

U: 4.0 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0 80.0

F: 0.5 2.0 2.0 2.0 3.0 6.0 7.0 8.0 8.0 9.6 9.8 12.0 15.0 19.0 22.0

- Now find the mid value for both sets:

Median term ( U ) = ( n + 1 ) / 2

= ( 11 + 1 ) / 2 = 12/2 = 6th term

Median ( U ), 6th term = 17.0

Median term ( F ) = ( n + 1 ) / 2

= ( 15 + 1 ) / 2 = 16/2 = 8th term

Median ( F ), 8th term = 8.0

- We will now trim the smallest and largest observation from each set.

- In set ( U ) we see that smallest data corresponds to ( 4.0 ) while the largest data corresponds to ( 80.0 ). We will exclude these two terms and the trimmed set is defined as:

U': 5.0 5.0 6.0 11.0 17.0 18.0 23.0 33.0 35.0

- In set ( F ) we see that the smallest data corresponds to ( 0.5 ) while the largest data corresponds to ( 22.0 ). We will exclude these two terms and the trimmed set is defined as:

F': 2.0 15.0 12.0 8.0 8.0 7.0 6.0 19.0 3.0 9.8 9.6 2.0 2.0

- We will again use the previous formula to calculate means of trimmed samples ( U' ) and ( F' ) as follows:

$E ( U' ) = \frac{5.0+ 5.0+ 6.0+ 11.0+ 17.0+ 18.0+ 23.0+ 33.0+ 35.0}{9} \\\\E ( U' ) = 17$

$E ( F' ) = \frac{2.0 +2.0+ 2.0 +3.0+ 6.0+ 7.0+ 8.0+ 8.0+ 9.6+ 9.8+ 12.0+ 15.0+ 19.0}{13} \\\\E ( F' ) = 7.95384$

- The trimming percentage is known as the amount of data removed from the original sample from top and bottom of sample size of 11 and 15, respectively.

- We removed the smallest and largest value from each set. Hence, a single value was removed from both top and bottom of each data set. We can express the trimming percentage for each set as follows:

$T ( U ) = \frac{1}{11} * 100 = 9.091\\\\T ( F ) = \frac{1}{15} * 100 = 6.667$ %

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sesenic [268]

Answer:

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Step-by-step explanation:

Domain is the set of x-values (x axis) and Range is the set of y-values (y axis).

Now if you look at the relation (points given), you can see the 5 points corresponds to 1,2,3,4, adn 5 in the x axis (minutes). So this is the domain - 1,2,3,4,5.

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