Answer:
1) (2.2, -1.4)
2) (1.33, 1)
Detailed solution:
Question 1)
We are provided with two linear equations representing lines, and we need to find the intersection point that solves the system.
The lines given are:
Line 1 equation:
This line passes through points (0, 2.5) and (2.2, -1.4).
Line 2 equation:
The second line goes through (0, -3) and (2.2, -1.4).
According to the graph and data, the solution to the system is the coordinate where both lines intersect.
The solution to a system of linear equations is the coordinate pair common to both lines, i.e., the intersection point.
Here, both lines share the point (2.2, -1.4), indicating it is their intersection and the solution.
Therefore, the solution for question 1 is (2.2, -1.4).
Question 2)
The equations given are:
y = 1.5x - 1 Equation 1
y = 1 Equation 2
The method of substitution can be used to find the solution.
Replacing y from Equation 2 into Equation 1 gives:
1 = 1.5x - 1
Add 1 to both sides:
2 = 1.5x
Dividing both sides by 1.5 yields:
x = 2/1.5
x = 1.33
y = 1
Thus, the solution of the system is (1.33, 1).
Let Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be denoted as J, C, G, M, E, D, A, and S respectively. In part IV, we need to identify the pairs of potential clients that could potentially be selected. The sample space consists of all possible outcomes, therefore we create a set of all valid pairs, listed as follows: {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S)}. We can verify the number of elements in the sample space, n(S) is 1+2+3+4+5+6+7=28. This gives us the answer to the first question: What is the count of pairs of potential clients that can be randomly selected from the pool of eight candidates? (Answer: 28.) II) What is the chance of a certain pair being chosen? The chance of picking a specific pair is 1/28, as there’s just one way to select a particular pair out of the 28 possible options. III) What is the probability that the selected pair consists of Jacob and Meg or Geraldo and Sally? The probability of selecting (J, M) or (G, S) is 2 out of 28, which equates to 1/14. Answers: I) 28 II) 1/28 ≈ 0.0357 III) 1/14 ≈ 0.0714 IV) {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S).}
The average years of employment is 14
with 73% having worked for at least 10 years.
To find the mean, add up the years of service and divide by the number of employees. The total years worked is 417, so the formula yields:
average years worked = 417/30 = 13.9, rounded to approximately 14 years.
To determine the percentage of employees with ten or more years, count those with 10 or more years and divide by the total employee count, converting the result into a percentage:
(10 years or over)/(total number) = 22/30 = 0.73 repeating, which approximates to 73%.
Using a spreadsheet tool can make this calculation simpler, as it can quickly compute the average and help tally how many employees have worked for 10 years or more.
