How can properties of operations be used to solve problems involving integers and rational numbers ​

(a) x= 0,1,2..........10 (b) p=0.02 (c) standard deviation = 0.44 (d) p=0.18. A comprehensive explanation of the answer is available in the attached file.

Answer:

A) Maia had 1920 beads more.

B) Maia had 2544 beads at first.

Step-by-step analysis:

Let x denote the beads with Jenny and y for Maia.

The information provided states that Maia possesses 2064 beads more than Jenny, which can be represented mathematically as:

Additionally, after Maia made a necklace with 144 beads, she had five times more beads than Jenny.

This can also be formulated as an equation:

A) Since Maia had 2064 beads more than Jenny before using 144 beads, we calculate her final bead count by subtracting 144 from 2064.

Thus, Maia is left with 1920 beads more than Jenny.

B) To solve this system of linear equations, we will utilize the substitution method.

By inserting equation (1) into equation (2), we arrive at:

Now, let's divide our equation by 4.

Now, we'll substitute x=480 into equation (1) to isolate y.

Conclusively, Maia initially had 2544 beads.

The circumcenter of the triangle is at the coordinates (2,1). To find this, we can outline the triangle based on the coordinates of points A (-1,5), B (-1,-3), and C (5,-3). Utilizing the distance formula helps us confirm that the triangle satisfies Pythagoras' theorem, hence it is a right triangle. In a right triangle, the circumcenter is located at the midpoint of the hypotenuse, which is determined to be (2,1).

114/6=9 parts per minute. 9 parts * 15 minutes= 135 parts

Answer:

Option 1 is valid, which entails 2 hours of walking and 12 hours of running.

Step-by-step explanation:

The equations provided are:

3w + 6r ≥ 36

3w + 6r ≤ 90

We'll assess which options comply with these equations.

1) 2 hours walking; 12 hours running

w = 2 and r = 12

3w + 6r ≥ 36

3(2) + 6(12) ≥ 36

6+72 ≥ 36

78 ≥ 36

3w + 6r ≤ 90

3(2) + 6(12) ≤ 90

6+72 ≤ 90

78 ≤ 90

Both equations are satisfied. Option 1 is valid.

2) 4 hours walking; 3 hours running

w = 4 and r = 3

3w + 6r ≥ 36

3(4) + 6(3) ≥ 36

12+18 ≥ 36

30 ≥ 36 (this does not hold since 30 < 36)

3w + 6r ≤ 90

3(4) + 6(3) ≤ 90

12+18 ≤ 90

30 ≤ 90

Thus, Option 2 is invalid.

3) 9 hours running; 12 hours walking

w = 9 and r = 12

3w + 6r ≥ 36

3(9) + 6(12) ≥ 36

27+72 ≥ 36

99 ≥ 36

3w + 6r ≤ 90

3(9) + 6(12) ≤ 90

27+72 ≤ 90

99 ≤ 90 (this does not hold since 99 > 90)

Option 3 is invalid.

4) 12 hours walking; 10 hours running

w = 12 and r = 10

3w + 6r ≥ 36

3(12) + 6(10) ≥ 36

36+60 ≥ 36

96 ≥ 36

3w + 6r ≤ 90

3(12) + 6(10) ≤ 90

36 + 60 ≤ 90

96 ≤ 90 (this does not hold since 96 > 90)

So, Option 4 is invalid.