5. A construction area has 6 cylindrical posts protecting the

Tickets to a school production cost $5 for a student ticket and $10 for an adult ticket. A total of 67 tickets were purchased at

vazorg [7]

Answer:

Step-by-step explanation:

Keywords:

System of equations, variables, cost, tickets, adults, children.

For this case we must solve a system of equations with two variables represented by the tickets of students and adults of a school production.

We define the variables according to the given table:

a: Number of tickets sold to adults

c: Amount of tickets sold to children.

We then have the following system of equations:

A + c = 67

10a + 5c =440

From the first equation, we clear the value of the variable c:

C = 67 - a

Answer:

The value that could replace c in the table is:

C = 67 - a

Option C is the answer!

Hope it helped u if yes mark me BRAINLIEST!

Tysm! Plz

9 0

3 years ago

Read 2 more answers

There are 42 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time

harkovskaia [24]

Answer:

A) 0.99413

B) 0.00022

Step-by-step explanation:

A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:

Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes

Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:

μ = n*μ_s ample = 42 × 5 = 210 minutes

While the standard deviation for the population would be:

σ = √nσ_sample = √(42 × 6) = 15.8745 minutes

To find the z-score, we will use the formula;

z = (x - μ)/σ

Thus;

z = (250 - 210)/15.8745

z = 2.52

From the z-distribution table attached, we have;

P(Z < 2.52) ≈ 0.99413

B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.

Thus, total time is now 250 + 10 = 260 minutes

Similar to the z-formula in A above, we have;

z = (260 - 210)/15.8745

z = 3.15

P(Z > 3.15) = 0.00022

5 0

2 years ago

Solve: m-9=3(1/4m+4)-20

FrozenT [24]

Answer: M = -20

Too easy

8 0

3 years ago

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves

Radda [10]

A washer has outer radius $2-x^2$ and inner radius $2 - \sqrt{x}$

$\displaystyle V = \int_0^1 \pi\Big[ \left( 2 - x^2 \right)^2 - \left( 2 - \sqrt{x}\right)^2\Big] dx$

4 0

3 years ago

Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow

Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

$y=\frac{x^2}{2}-3$ with $x \in \{-4,-2,0,2}$ .

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

$y=\frac{x^2}{2}-3$ with $x=-4$ :

$y=\frac{(-4)^2}{2}-3$

$y=\frac{16}{2}-3$

$y=8-3$

$y=5$ .

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

$y=\frac{x^2}{2}-3$ with $x=-2$ :

$y=\frac{(-2)^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1. Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

$y=\frac{x^2}{2}-3$ with $x=0$ :

$y=\frac{0^2}{2}-3$

$y=\frac{0}{2}-3$

$y=0-3$

$y=-3$

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3. Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

$y=\frac{x^2}{2}-3$ with $x=2$

$y=\frac{2^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed. The graph shows this point as well.

13) The answer given is correct.

$g(x)=|x|$ is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

$m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|$ .

If you move that up once it becomes:

$n(x)=|x-3|+1$ or $n(x)=|3-x|+1$ which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is. That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

$f(7)=|3-7|+1=|-4|+1=4+1=5$

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

8 0

2 years ago