(b) if the length, width, and height are increasing at a rate of 0.1 ft/sec, 0.2 ft/sec, and 0.5 ft/sec respectively, find the r

<span>5.7 liters of a 5% solution combined with 4.3 liters of a 40% solution. To begin, define the problem with formulas. x Represents the liters of the 5% solution utilized. 10-x Represents the liters of the 40% solution used. This forms an equation: 5% of x plus 40% of (10-x) equals 20% of 10. 0.05x + 0.40(10-x) = 0.20 * 10 Now, distribute the 0.40 coefficient. 0.05x + 4.0 - 0.40x = 0.20 * 10 Next, combine the terms. 4.0 - 0.35x = 2.0 Add 0.35x to each side. 4.0 = 2.0 + 0.35x Subtract 2 from both sides. 2.0 = 0.35x Lastly, divide both sides by 0.35. 5.7 = x Thus, 5.7 liters of a 5% solution is required. To determine the volume of the 40% solution, subtract from 10. 10.0 - 5.7 = 4.3</span>

1.4×5=7

0.8×10=8

1.4×10=14

1×15=15

15+14+8+7=44

44÷4=11

LQ of 44=11

LQ=10 minutes

11×3=33

UQ= 29 minutes

The Range is 19 minutes

Detailed breakdown:

Commence with the individual boxes. For determining the number of students in each category, calculate Frequency density × The difference in the category. (if it's 5-15, the difference is 10)

This results in the counts of students in each range.

Next, determine the LQ of 44, which is 11.

Then locate the 11th student's score; in this instance, it resides in the 5-15 range. 7 students have already surpassed it, with 8 in the 5-15 range. Hence, the 11th lies within the bounds of 5-15, making the middle 10.

Repeat this process for the UQ.

The interquartile range is calculated as UQ-LQ, yielding 29-10=19 minutes.

I hope this helps, though I'm not entirely sure if my explanation is coherent and I'm unclear on the terminology I've used for these categories.

<span>P(black socks): 24/42 or 12/21
P(black socks without replacement): 23/41. Consequently, the chance of randomly selecting 2 black socks, without replacement, from the basket is 12/21×23/41=276/861, equivalent to 32%. Hope this helps!</span>

These occurrences are unlikely to take place!

I hope this is helpful

Answer:

Step-by-step explanation:

Given that lines l and m are parallel and there is a transversal cutting through these lines.

5). (9x + 2)° = 119° [by alternate interior angles]

    9x = 117 ⇒ x = 13

6). (12x - 8)° + 104° = 180°

     12x = 180 - 96

     x = ⇒ x = 7

7). (5x + 7) = (8x - 71) [by alternate exterior angles]

    8x - 5x = 71 + 7

    3x = 78

     x = 26

8). (4x - 7) = (7x - 61) [by corresponding angles]

    7x - 4x = -7 + 61

    3x = 54

     x = 18

9). (9x + 25) = (13x - 19) [by corresponding angles]

    13x - 9x = 25 + 19

    4x = 44

     x = 11

   (13x - 19)° + (17y + 5)° = 180° [linear pairs of angles sum to supplementary]

    (13×11) - 19 + 17y + 5 = 180

    129 + 17y = 180

    17y = 180 - 129

     y = 3

10). (3x - 29) + (8y + 17) = 180 [linear pairs of angles sum to supplementary]

     3x + 8y = 180 + 12

     3x + 8y = 192 -----(1)

     (8y + 17) = (6x - 7) [by alternate exterior angles]

     6x - 8y = 24

     3x - 4y = 12 -----(2)

     From equation (1) subtract equation (2)

     (3x + 8y) - (3x - 4y) = 192 - 12

     12y = 180

     y = 15

     Plugging into equation (1),

     3x + 8(15) = 192

     3x + 120 = 192

    x = 24

11). (3x + 49)° = (7x - 23)° [by corresponding angles]

    7x - 3x = 49 + 23

    4x = 72 ⇒ x = 18

    (11y - 1)° = (3x)° [by corresponding angles]

    11y = 3×18 + 1

     11y = 55 ⇒ y = 5

12). (5x - 38)° = (3x - 4)° [by corresponding angles]

     5x - 3x = 38 - 4

     2x = 34

     x = 17

     (7y - 20)° + (5x - 38)° + 90° = 180°

     [The angles within a triangle sum to 180°]

     7y + 5x - 58 = 90

     5x + 7y = 148

     5×17 + 7y = 148

     85 + 7y = 148

     7y = 148 - 85

     y =