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3 years ago

HURRY 12 POINTS

Mathematics

1 answer:

Svetlanka [38]3 years ago

8 0

Answer:

The correct description of the variables is given by as the following -

The variables are gender and preferred method of learning. They are categorical variables.

Step-by-step explanation:

The correct description of the variables is given by as the following -

The variables are gender and preferred method of learning. They are categorical variables.

You might be interested in

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

If there are 37 figures and there are six people how many will each person get

Elis [28]

Answer:

It depends.

Step-by-step explanation:

If we're able to brake the figures, then each person will get six and one sixth of a figure

$37 \div 6 = 6 \frac{1}{6}$

But if we can't break them, then each person will get six figures with one remaining ungiven.

$37 \div 6 = 6 \: (1)$

5 0

3 years ago

Nvm I got it now lol

Usimov [2.4K]

Answer:

nice bro good for you :)

Step-by-step explanation:

3 0

2 years ago

A company manufactures two different sizes of boat lifts. The smaller lift requires 1 hour in the welding department and 2 hours

qaws [65]

Answer:

The solution that optimizes the profit is producing 0 small lifts and 50 large lifts.

Below are all the steps explained in detail.

The graph is attached.

Explanation:

1. Name the variables:

x: number of smaller lifts
y: number of larger lifts

2. Build a table to determine the number of hours each lift requires from each department:

Number of hours

small lift large lift total per department

Welding department 1x 3y x + 3y

Packaging department 2x 1y 2x + y

3. Constraints

150 hours available in welding: x + 3y ≤ 150
120 hours available in packaging: 2x + y ≤ 120
The variables cannot be negative: x ≥ 0, and y ≥ 0

Then you must:

draw the lines and regions defined by each constraint
determine the region of solution that satisfies all the constraints
determine the vertices of the solution region
test the profit function for each of the vertices. The vertex that gives the greatest profit is the solution (the number of each tupe that should be produced to maximize profits)

4. Graph

See the graph attached.

Here is how you draw it.

x + 3y ≤ 150
draw the line x + 3y = 150 (a solid line because it is included in the solution set)
shade the region below and to the left of the line

2x + y ≤ 120
draw the line 2x + y ≤ 120 (a solid line because it is included in the solution set)
shade the region below and to the left of the line

x ≥ 0 and y ≥ 0: means that only the first quadrant is considered

the solution region is the intersection of the regions described above.

take the points that are vertices inside the solutoin region.

5. Test the profit function for each vertex

The profit function is P(x,y) = 25x + 90y

The vertices shown in the graph are:

(0,0)
(0,50)
(42,36)
(60,0)

The profits with the vertices are:

P(0,0) = 0
P(0,50) = 25(0) + 90(50) = 4,500
P(42,36) = 25(42) + 90(36) = 4,290
P(60,0) = 25(60) + 90(0) = 1,500

Thus, the solution that optimizes the profit is producing 0 smaller lifts and 90 larger lifts.

3 0

2 years ago

The price of tomatoes went from $1.12 per pound to $1.96 per pound in 3 years. find the rate of change of the price of tomatoes.

ad-work [718]

$\bf \begin{array}{|cc|ll} \cline{1-2} \stackrel{x}{year}&\stackrel{y}{price}\\ \cline{1-2} 0&1.12\\ 3&1.96\\ \cline{1-2} \end{array}~\hspace{10em} (\stackrel{x_1}{0}~,~\stackrel{y_1}{1.12})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1.96}) \\\\\\ \stackrel{\textit{rate of change}}{slope = m\implies} \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1.96-1.12}{3-0}\implies \cfrac{0.84}{3}\implies 0.28$

7 0

3 years ago