Antonio wants to examine the sales in a restaurant during a period of time from 2 hours before to 2 hours after the peak dinner time and graph his results. If the peak dinner time is considered to be at 6 p.m., what would be the domain for the graph Antonio is creating?

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

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)

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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:

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:

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:

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]

Answer:

4,-1

Step-by-step explanation:

Please like!♥

We know that:

Now, let's simplify.

Therefore, Anita will need to visit Brookshire 30 more times to receive a free set of dishes.

Hoped this helped

[tex]BrainiacUser1357[/tex]

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.

Answer:

Step-by-step explanation:

y = (4/7)x - 1

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: 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:

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

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.

If any doubt leave a comment

Answer:

6/6 or 1

Step-by-step explanation:

bow tie method

Answer:

D

Step-by-step explanation:

:)

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

Answer:

2x^2 + 11x – 1

Step-by-step explanation:

Answer:

2x^3

C

Step-by-step explanation:

Answer:

[tex]r \geqslant 10.4[/tex]

The maximum rate at which the diver will descend will be 7 feet per hour.

The rate of change is the change of a quantity over 1 unit of another quantity.

Most of the time the rate of change is the change with respect to time.

For example the speed 3meter/second.

As per the given,

Initial depth of scuba diver = 12.8 feet

Let's say the rate of a scuba diver is "r".

Initial depth + depth cover with constant rate ≤ final depth

12.8 + (312/60)r ≤ 49.2

r ≤ 7

Thus, the maximum rate at which a diver can descend is 7 feet per hour.

Hence"The maximum rate at which the diver will descend will be 7 feet per hour".

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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