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MEN11 11/15/12 Mrs. Cabanero Goal: To be able to solve problems involving *percent increase or decrease and. *percent error Do Now: Use the given formula to solve the ff : % increase(or decrease) =
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MEN11 11/15/12 Mrs. CabaneroGoal: To be able to solve problems involving *percent increase or decrease and *percent error Do Now: Use the given formula to solve the ff: % increase(or decrease) = *The world population was 4.2 billion people in 1982. The population in 1999 reached 6 billion. Find the percent of increase from 1982 to 1999.
A percent increase or decrease gives the ratio of the amount of increase or decrease to the original amount. Error is the absolute value of the difference between a measured value and the true value. Relative error is the ratio of error to the true value. relative error =
Regents: At the end of week one, a stock had increased in value from $5.75 a share to $7.50 a share. Find the percent of increase at the end of week one to the nearest tenth of a percent. At the end of week two, the same stock had decreased in value from $7.50 to $5.75. Is the percent of decrease at the end of week two the same as the percent of increase at the end of week one? Justify your answer.
Part1. week onePercent increase = = Part2. week two Percent decrease = Part3. Therefore, the percent of increase is not the same as the percent of decrease.
REGENTSCarrie bought new carpet for her living room. She calculated the area of the living room to be 174.2 square feet. The actual area was 149.6 square feet. What is the relative error of the area to the nearest ten-thousandth?
Step 1. Given: calculated area = 174.2 sq.feet actual area = 149.6 sq. feet Step2. answer
REGENTSThe actual dimensions of a rectangle are 2.6 cm by 6.9 cm. Andy measures the sides as 2.5 cm by 6.8 cm. In calculating the area, what is the relative error, to the nearest thousandth?
Step1. actual area = 2.6 sq. cm measured area = 2.5 Step2. relative error = 0.0523 = 0.052 answer
EXIT TICKET: to be collected The dimensions of a rectangle are measured to be 12.2 inches by 11.8 inches. The actual dimensions are 12.3 inches by 11.9 inches. What is the relative error, to the nearest ten-thousandth, in calculating the area of the rectangle?