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3 years ago

Which section of the graph represents the solution of the system of inequalities shown?

Mathematics

2 answers:

faltersainse [42]3 years ago

4 0

<h3>Answer: Region Q</h3>

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Explanation:

Let's focus on the inequality 4x-2y > 6

Plug in (x,y) = (0,0) to find that...

4x-2y > 6

4(0)-2(0) > 6

0 > 6

This is a false statement. So that means (0,0) is <u>not</u> in the shaded region for 4x-2y > 6. So we'll shade the opposite side of the dashed line to shade regions Q and R (i.e. stuff below the dashed line).

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Let's check the point (0,0) with the other inequality as well

$5x + 15y \ge -20\\\\5(0) + 15(0) \ge -20\\\\0 \ge -20\\\\$

This is true because 0 is to the right of -20 on the number line.

So we'll shade regions P and Q to represent the solution set for this inequality. These regions are above the boundary line. Points on the boundary are also included.

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To summarize so far, we found that

regions Q and R make 4x-2y > 6 true,
regions P and Q make $5x + 15y \ge -20\\\\$ true.

The overlap is region Q which is the final answer

Any point from region Q satisfies <u>both</u> 4x-2y > 6 and $5x + 15y \ge -20\\\\$ at the same time. A point on the solid boundary line is part of the solution set, but stuff on the dashed boundary line are not solution points.

3241004551 [841]3 years ago

3 0

Answer:

Step-by-step explanation:

5x+15y ≥ -20

Solve for y

15 y≥ -5x-20

Divide by 15

y ≥ -5x/15 -20/15

y ≥ -1/3 x -4/3

Y is greater than or equal to so shade above and on the line

4x-2y > 6

Solve for y

-2y > -4x+6

y < 2x -3

Shade below the line

The double shaded area is the solution

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Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The following ar

Softa [21]

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:

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

- The trimming pecentages for each data set are 9.091% and 6.667% respectively.

7 0

4 years ago

Sum of two number is 120. If the larger number is four times the smaller number, wbat are the two numbers

tankabanditka [31]

x+y=120_______1

x=4*y_________2

x=4y

substitute 4y for x in equation 1

`4y+y=120

5y=120

eliminate the coefficient of y

y=120/5

y=24

now,input 24 for y in equation 2

`x=4y

x=4*24

x=96

thus, the numbers are

96 and 24

3 0

3 years ago

The variable k represents a value in the set {-2-4-6-8}. which value of k makes 4 (1.5-k)-10=56/2 a true statement A.-2 B. -4 C.

patriot [66]

Question: The variable k represents a value in the set {-2,-4,-6,-8}. Which value of k make 4(1.5-k)-10=56/2 a true statement?

Step-by-step explanation:

brainly.com/question/23369106

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3 years ago

A car company claims that its cars achieve an average gas mileage of at least 26 miles per gallon. A random sample of eight cars

motikmotik

Answer:

$t=\frac{25.5-26}{\frac{1}{\sqrt{8}}}=-1.414$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value for this case is given by:

$p_v =P(t_{(7)}$

Since the p value is higher than the significance level of 0.06 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 25.5 and then the claim makes sense

Step-by-step explanation:

Information given

$\bar X=25.5$ represent the sample mean