HELP PLEASE!!!!
Which of the following statements are true about the data set represented by this box plot? Select all that apply.

The interquartile range is 26.
The maximum value is 44.
The range is 50.
The median is 25.

Answer:

104in2

Step-by-step explanation:

The area, in square inches, of the painted face of the brick is Option C 104 [tex]in^{2}[/tex].

Using the Quadrant system of the Number line

B (-4,2) lies in the 2nd quadrant.

C (-4,-11) lies in the 3rd quadrant

A (4,2) lies in the 1st quadrant

D (4,-11) lies in the 4th quadrant

Length of Side AB= Length of Side CD= Distance between x coordinate of A & B OR Between C & D i.e. 8 in

Length of Side BC= Length of side  AD= Distance between y coordinate of A & D OR Between B & C i.e. 13 in

Area of the Square ABCD= (length of Side AB or CD) X (Length of Side  BC or AD)

                                          = (8 x 13) [tex]in^{2}[/tex]

                                          = 104 [tex]in^{2}[/tex]

Thus, Area of square is 104 [tex]in^{2}[/tex].

To learn more about area: https://brainly.com/question/25292087

Answer:

77.5

Step-by-step explanation:

Given:

Scores: 65, 70, 75, 80, 85, 90, 95

Number of Students: 8, 5, 3, 6, 6, 4, 2

Step 1: Calculate the sum

Sum = (65 * 8) + (70 * 5) + (75 * 3) + (80 * 6) + (85 * 6) + (90 * 4) + (95 * 2)

= 520 + 350 + 225 + 480 + 510 + 360 + 190

= 2635

Step 2: Calculate the mean

Mean = Sum / Total Number of Students

= 2635 / (8 + 5 + 3 + 6 + 6 + 4 + 2)

= 2635 / 34

≈ 77.5

Answer:77.5

Step-by-step explanation:

The chemist needs to use 5 liters of the 30% solution and 5 liters of water.

Answer:

30% solution x liters

30%(water) 10-x liters

15% 10 liters

Step-by-step explanation:

In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex],  Answer 5th is 384 [tex]in^{2}[/tex],  Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.

2) In the 2nd figure, we have 5 faces as two equal Right angle Triangles (on two opposite faces) and three rectangles(on three sides of triangles)

Dimensions of both Right Angle Triangle:

Perpendicular = 9in

Base = 12 in ;

Dimensions of Rectangles:

Rectangle 1 : L1 = 15 in

                    B1 = 3 in

Rectangle 2: L2 = 9 in

                    B2 = 3 in

Rectangle 3: L3 = 12 in

                    B3 = 3 in

Area of 2nd figure = Sum of the area of its 5 faces

= Area of two Right angle Triangles + Area of Rectangle 1 + Area of rectangle 2 + Area of Rectangle 3

= [tex]\frac{1}{2}[/tex] * perpendicular * Base + L1*B1 + L2*B2 + L3*B3   ...................... (Formulas for particular Figures)

= [tex]\frac{1}{2}[/tex]* 9*12 + 15*3 + 9*3 + 12*3 [tex]in^{2}[/tex]

= 54+ 45+ 27 + 36 [tex]in^{2}[/tex]

=162 [tex]in^{2}[/tex]

Area of 2nd figure = 162 [tex]in^{2}[/tex]

5) In the 5th figure, we have a cube with all sides equals to 8in. Cube has 6 faces.

The area of the figure is = [tex]6(side)^{2}[/tex]

= [tex]6(8)^{2}[/tex][tex]in^{2}[/tex]

=384 [tex]in^{2}[/tex]

Area of 5th figure = 384 [tex]in^{2}[/tex]

6) Total Amount of wrapping paper needed is the sum of the Area of all Figures

= 364 + 162 + 290 + 192+ 384 [tex]in^{2}[/tex]

= 1392 [tex]in^{2}[/tex]

The total amount of wrapping paper needed is = 1392 [tex]in^{2}[/tex]

7) One Roll of wrapping paper with measurements 30 in and 60 in has an Area

= 30*60 [tex]in^{2}[/tex]

= 1800 [tex]in^{2}[/tex]

As we need only 1392 [tex]in^{2}[/tex] of wrapping paper which is less than one roll of wrapping paper so only 1 roll is required

hence, In this Drag and Drop Activity Answer 2nd is 162 [tex]in^{2}[/tex],  Answer 5th is 384 [tex]in^{2}[/tex],  Answer 6th is 1392 [tex]in^{2}[/tex] , Answer 7th is 1 roll.

To learn more about Area of Rectangle

https://brainly.com/question/2607596

Answer:

In explanation

Step-by-step explanation:

To solve the rational equation 1/x + 1/(x + 2) = 9/40, we can start by finding a common denominator and then simplifying the equation. The common denominator for the two fractions is (x)(x + 2). Multiplying each term by this common denominator, we get:

[(x + 2) + x] / (x)(x + 2) = 9/40

Simplifying the numerator, we have:

(2x + 2) / (x)(x + 2) = 9/40

Now, we can cross-multiply to eliminate the denominators:

40(2x + 2) = 9(x)(x + 2)

Expanding both sides, we have:

80x + 80 = 9x^2 + 18x

Rearranging the equation and setting it equal to zero:

9x^2 - 62x - 80 = 0

Now, we can factor the quadratic equation:

(9x + 10)(x - 8) = 0

Setting each factor equal to zero, we have:

9x + 10 = 0 or x - 8 = 0

Solving for x in each case:

9x = -10 or x = 8

Dividing both sides of the first equation by 9, we find:

x = -10/9 or x = 8

Since we are looking for even integers, we can disregard the negative solution. Therefore, the value of x is 8.

Hence, the two consecutive even integers can be represented by x and x + 2, which gives us 8 and 10. So, the integers are 8 and 10.

Answer:

He forgot to flip the second fraction.

Step-by-step explanation:

He has:

−8/9 ÷ 9/8 = −8/9 × 9/8 = −72/72 = −1

He made a mistake. When you change a division to a multiplication, you must use the reciprocal of the second fraction. You have to flipt the second fraction. He did not flip it.

This is correct:

−8/9 ÷ 9/8 = −8/9 × 8/9 = −64/81

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

Reason:

This is a geometric series with

The template is [tex]a(r)^{n-1}[/tex]

If -1 < r < 1, then the infinite geometric series converges to a finite number. This is because we add on smaller and smaller pieces, which prevents the sum going off to infinity.

In the case of r = -1/3, it fits the interval  -1 < r < 1. In other words -1 < -1/3 < 1 is true.

We'll plug those values into the formula below to wrap things up.

[tex]S = \frac{a}{1-r}\\\\S = \frac{-4}{1-(-1/3)}\\\\S = \frac{-4}{1+1/3}\\\\S = \frac{-4}{3/3+1/3}\\\\[/tex]

[tex]S = \frac{-4}{4/3}\\\\S = -4 \div \frac{4}{3}\\\\S = \frac{-4}{1} \times \frac{3}{4}\\\\S = \frac{-4*3}{1*4}\\\\S = -3\\\\[/tex]

Therefore,

[tex]\displaystyle \sum_{n=1}^{\infty} -4\left(-\frac{1}{3}\right)^{n-1} = -3[/tex]

The final answer is -3.

You can verify the answer by generating partial sums with a spreadsheet. The partial sums should steadily get closer to -3.

Here's a few partial sums.

[tex]\begin{array}{|c|c|c|} \cline{1-3}\text{n} & \text{a}_{\text{n}} & \text{S}_{\text{n}}\\\cline{1-3}1 & -4 & -4\\\cline{1-3}2 & 1.333333 & -2.666667\\\cline{1-3}3 & -0.444444 & -3.111111\\\cline{1-3}4 & 0.148148 & -2.962963\\\cline{1-3}5 & -0.049383 & -3.012346\\\cline{1-3}6 & 0.016461 & -2.995885\\\cline{1-3}7 & -0.005487 & -3.001372\\\cline{1-3}8 & 0.001829 & -2.999543\\\cline{1-3}9 & -0.00061 & -3.000153\\\cline{1-3}10 & 0.000203 & -2.99995\\\cline{1-3}\end{array}[/tex]

The interesting thing is that the partial sums [tex]S_n[/tex]  bounce around -3 while also getting closer to it.

Answer:

-3/4

Step-by-step explanation:

-1 1/4 + 1/2 = -5/4 + 1/2 = -5/4 + 2/4 = -3/4

Answer:

4/7

Step-by-step explanation:

class total = 21

chance of picking one of 12 boys out of 21.

12/21 which can be simpilfied to 4/7