For the question the answer is a
1.50x + 6.50y is your algebraic expression.
For this case we have the following equation:

Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:

Making the corresponding calculation:
Answer:
the approximate weight of the spring in pounds is:
Given that the distance between two lines in the measurement instrument is 0.01m, the maximum error should be 0.01m/2 = 0.005 m.
If you do a good work, the real measure is 1.20m +/- 0.005m, this is between 1.195 and 1.205.