Question

1. find the c.o.p. of this table:

weeks assignments
1 7
2 14
3 21
4 25

Explanation:

Step1: Recall C.O.P. formula

C.O.P. (Constant of Proportionality) is \( \frac{y}{x} \) for proportional relationships, where \( y \) is the dependent variable and \( x \) is the independent variable. Here, \( x \) is Weeks, \( y \) is Assignments.

Step2: Check proportionality for first three pairs

• For \( x = 1, y = 7 \): \( \frac{7}{1} = 7 \)
• For \( x = 2, y = 14 \): \( \frac{14}{2} = 7 \)
• For \( x = 3, y = 21 \): \( \frac{21}{3} = 7 \)
• For \( x = 4, y = 25 \): \( \frac{25}{4} = 6.25 \) (not 7, so only first three are proportional)

Step3: Determine C.O.P.

Since first three pairs have \( \frac{y}{x}=7 \), the C.O.P. for the proportional part is 7. (Note: The 4th pair breaks proportionality, but C.O.P. is for proportional relationships, so we use the consistent ratio.)

Answer:

7 (for the proportional part of the table, weeks 1 - 3)