Which point represents the outlier

Answer:

6/6 or 1

Step-by-step explanation:

bow tie method

Check the picture below.

Answer:

2x^3

C

Step-by-step explanation:

The vertices of the feasible region of the constraints is (0,2.5)

The constraints are given as:

x + 3y ≤ 6

4x + 6y ≥ 9

x ≥ 0,  y ≥ 0

Express the inequalities as equations

x + 3y = 6

4x + 6y = 9

Make x the subject in x + 3y = 6

x = 6 - 3y

Substitute x = 6 - 3y in 4x + 6y = 9

4(6 - 3y) + 6y = 9

Expand

24 - 12y + 6y = 9

Evaluate the like terms

-6y = -15

Divide by 6

y = 2.5

Substitute y = 2.5 in x = 6 - 3y

x = 6 - 3 * 2.5

Evaluate

x = -1.5

So, we have:

(x,y) = (-1.5, 2.5)

Recall that: x ≥ 0,  y ≥ 0

This means that the feasible region must have positive coordinates or zero

So, we set x = 0

(x,y) = (0, 2.5)

Hence, the vertices of the feasible region of the constraints is (0,2.5)

Read more about feasible region at:

https://brainly.com/question/14381991

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If any doubt leave a comment

Answer:

4,-1

Step-by-step explanation:

Please like!♥

Explanation :

________________________________

Given:

(2+√3)+(4-√3)

To Find:

Evaluate the expression

Solution:

⇨Step 1:

Remove the parentheses

(2+√3)+(4-√3)

When there is + or no sign in front of an expression in parentheses, the expression remains the same

= 2 + √3+ 4-√3

⇨Step 2:

Cancel the opposite terms √3 and -√3

"Two numbers are opposites if they have the same absolute value but different signs"

= 2 + 4

⇨Step 3:

Add the numbers

= 6

Hence (2+√3)+(4-√3) = 6.

Answer:

D

Step-by-step explanation:

:)

Answer:

300/75

Step-by-step explanation:

Cost of the Car = $15,000

a) Rate of interest received on savings(r) = 2%

Amount in savings account(p) =$3000

Time period(t) = 5 years    

simple intrest.

Interest received on savings account = $300

b) Rate of interest on loan = 0.5%

IF P= 15000

intrest.

Difference between the interest = $75

c) Taking loan of the whole amount that is of $15,000 is more reasonable because though the interest is more but Troy will receive interest($300) from his savings account as well. But if he withdraws $3000 from savings and takes the loan for the rest of the amount, he would have no earnings.

Answer: the answer is 7.5

Step-by-step explanation:

Answer:

Negative

Step-by-step explanation:

The slope of the line is negative because when looking at the graph, you can see that the line starts in the 2nd quadrant and ends in the 4th.

Answer:

15+20=35+15 And then add an additional plus 15 for each hour

Step-by-step explanation:

Answer:

0.0007361111111111111+0.001;

0.0027361111111111111-0.001;

1.0017361111111111111-1;

Step-by-step explanation:

you calculate 8^(-2)*3^(-2) that gives you 0.0017361111111111111 and then you add 1 anywhere you want and then say -1 where you put that +1 and that's all

Answer:

[tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Step-by-step explanation:

I'm assuming the system is [tex]\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.[/tex]:

[tex]x^2+y^2=2[/tex]

[tex]x^2+(2x^2-3)^2=2[/tex]

[tex]x^2+(4x^4-12x^2+9)=2[/tex]

[tex]x^2+4x^4-12x^2+9=2[/tex]

[tex]4x^4-11x^2+9=2[/tex]

[tex]4x^4-11x^2+7=0[/tex]

[tex]x^4-11x^2+28=0[/tex]

[tex](x^2-7)(x^2-4)=0[/tex]

[tex](4x^2-7)(x^2-1)=0[/tex]

[tex]4x^2-7=0[/tex]

[tex]4x^2=7[/tex]

[tex]x^2=\frac{7}{4}[/tex]

[tex]x=\pm\sqrt{\frac{7}{4}}[/tex]

[tex]x^2-1=0[/tex]

[tex]x^2=1[/tex]

[tex]x=\pm1[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm\sqrt{\frac{7}{4}})^2-3[/tex]

[tex]y=2({\frac{7}{4}})-3[/tex]

[tex]y=\frac{7}{2}-3[/tex]

[tex]y=\frac{1}{2}[/tex]

[tex]y=2x^2-3[/tex]

[tex]y=2(\pm1)^2-3[/tex]

[tex]y=2(1)-3[/tex]

[tex]y=2-3[/tex]

[tex]y=-1[/tex]

Therefore, [tex]x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}[/tex] and [tex]y=-1,\frac{1}{2}[/tex]

The ordered pair solutions are [tex](-\sqrt{\frac{7}{4}},0.5)[/tex], [tex](\sqrt{\frac{7}{4}},0.5)[/tex], [tex](-1,-1)[/tex], and [tex](1,-1)[/tex].

Answer:

this might help

Step-by-step explanation:

this might be kinda similar

Chickens and rabbits

In the yard were chickens and rabbits. Together they had 18 heads and 56 legs. How many chickens and how many rabbits were in the yard?

Correct answer:

chickens:  8

rabbits:  10

Step-by-step explanation:

a+b=18

2a+4b=56

a=8

b=10

a+b = 18

2•a+4•b=56

a+b = 18

2a+4b = 56

a = 8

b = 10

Answer:

y = 6x - 3

Step-by-step explanation:

Answer:

Step-by-step explanation:

1. False

2. False

3. True

4. True

5. True

6. True

7. True

8. True

9. True

Answer:

the answer you're looking for is D. when you move to the left the -4 becomes -6 and when you move up 2 becomes 4.

Let X = total cost of game

          (4/5) X = $25

Multiply both sides of equation by 5/3

   (5/3)(3/5) X = (5/3) $25

                   X = $35

Answer:

B

Step-by-step explanation:

The dot on 2 is an open spot, so 2 isn't part of the solution. the arrow points towards +∞ so, yeah. b

Using the t-distribution, it is found that since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.

At the null hypothesis, it is tested if isolated compounds do not give more counts than those from background, that is:

[tex]H_0: \mu_1 - \mu_2 \leq 0[/tex]

At the alternative hypothesis, it is tested if the give more counts, that is:

[tex]H_1: \mu_1 - \mu_2 > 0[/tex]

Using a calculator, the mean and the standard deviation for each sample are given by:

[tex]\mu_1 = 33.2, s_1 = 6.058[/tex]

[tex]\mu_2 = 24.25, s_2 = 4.35[/tex]

Considering the sample sizes of 5 and 4, respectively, the standard errors are given by:

[tex]s_1 = \frac{6.058}{\sqrt{5}} = 2.71[/tex]

[tex]s_2 = \frac{4.35}{\sqrt{4}} = 2.175[/tex]

For the distribution of differences, the mean and standard error are given by:

[tex]\overline{x} = \mu_1 - \mu_2 = 33.2 - 24.25 = 8.95[/tex]

[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.71^2 + 2.175^2} = 3.4749[/tex]

The standard error was found from the standard deviation for each sample, hence, the t-distribution is used.

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis, hence:

[tex]t = \frac{\overline{x} - \mu}{s}[/tex]

[tex]t = \frac{8.95 - 0}{3.4749}[/tex]

[tex]t = 2.58[/tex]

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with 5 + 4 - 2 = 7 df and a 0.05 significance level is of [tex]t^{\ast} = 1.895[/tex]

Since the test statistic is more than the critical value for the right-tailed test, it can be conclude that the isolated compound gives more counts than those from background, and hence, we can be 95% confident that the compound is indeed radioactive.

A similar problem is given at brainly.com/question/24826023

Answer:

D. x = negative eleven-halves plus-or-minus StartFraction 5 StartRoot 5 EndRoot Over 2 EndFraction

Step-by-step explanation:

Evaluate 0 and 1 (xe+ex)Dx

I'll do the first two parts to get you started.

=====================================================

Part (i)

We can show that the operation [tex]\nabla[/tex] is not commutative by picking two random and different real numbers for a,b. In other words, I'm using a counter-example.

Let's say we go for a = 1 and b = 2

[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\1 \nabla 2 = \frac{3*1+2}{5} - 1\\\\1 \nabla 2 = \frac{5}{5} - 1\\\\1 \nabla 2 = 1 - 1\\\\1 \nabla 2 = 0\\\\[/tex]

Now let's swap the values. We'll try a = 2 and b = 1

[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\2 \nabla 1 = \frac{3*2+1}{5} - 1\\\\2 \nabla 1 = \frac{7}{5} - 1\\\\2 \nabla 1 = 1.4 - 1\\\\2 \nabla 1 = 0.4\\\\[/tex]

We can see that [tex]1 \nabla 2[/tex] and [tex]2 \nabla 1[/tex] are not the same value. In general, [tex]a \nabla b \ne b \nabla a[/tex] Therefore, the operation [tex]\nabla[/tex] is not commutative.

The only time [tex]a \nabla b = b \nabla a\\\\[/tex] is true is when [tex]a = b[/tex], since,

[tex]\frac{3a+b}{5}-1 = \frac{3b+a}{5}-1\\\\\frac{3a+a}{5}-1 = \frac{3a+a}{5}-1\\\\\frac{4a}{5}-1 = \frac{4a}{5}-1\\\\[/tex]

It's when [tex]a \ne b[/tex] is where the operation becomes noncommutative.

Another way to arrive at the [tex]a = b[/tex] condition is to solve the original equation for either 'a' or b.

=====================================================

Part (ii)

Using the previous part as inspiration, we'll do a counter-example to show that the operation is not associative. Pick 3 random values for a,b,c. Here are the values I'll pick.

Then,

[tex]d = b \nabla c = \frac{3b+c}{5}-1 = \frac{3*2+3}{5}-1 = 0.8\\\\a \nabla (b \nabla c) = a \nabla d = \frac{3a+d}{5}-1 = \frac{3*1+0.8}{5}-1 = -0.24\\[/tex]

Next, we can say:

[tex]d = a \nabla b = \frac{3a+b}{5}-1 = \frac{3*1+2}{5}-1 = 0\\\\(a \nabla b) \nabla c = d \ \nabla c = \frac{3d+c}{5}-1 = \frac{3*0+3}{5}-1 = -0.4[/tex]

Let's compare the two outputs:

[tex]a \nabla (b \nabla c) = -0.24\\\\(a \nabla b) \nabla c = -0.4[/tex]

They don't match up, so [tex]a \nabla (b \nabla c) \ne (a \nabla b) \nabla c[/tex] when (a,b,c) = (1,2,3).

In general, the operation is not associative in R.

Answer:

P = -3, each line in the negative direction is 1 number

so you would have -1, -2, -3 , -4, etc.

P is on the -3 line.

See attached picture.

Step-by-step explanation:

Answer:

2=-12.2/(-6.1

-2=16/(-8)

-3= -2 2/5/4/5

3= 3 3/7 / 1 1/7

Step-by-step explanation:

Answer:

Domain: [0 , ∞)

Range: (-∞ , 4]

Step-by-step explanation:

For domain, look along the x axis.

The graph starts from x = 0 and it's ending point isn't given in the graph as it continues moving along the positive x axis.

Hence, it's ending point must be ∞

Domain of a function is:

[minimum point in x, maximum point in x]

=> [0, ∞)

"),(" is used for infinity.

"],[" for points that lie within the domain of the function.

For range, look along the y axis.

The graph starts at y = 4, and continues moving downwards till infinity.

Range:

[minimum in y, maximum in y]

=> (-∞, 4]