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Bears. During the summer, a bear’s heart rate is about 60 beats per minute, but it can drop to as low as 8 beats per minute during the winter. What is the percent of change in a bear’s heart rate from the summer rate to the winter low rate?. EXAMPLE 1. Finding a Percent of Decrease.
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Bears During the summer, a bear’s heart rate is about 60 beats per minute, but it can drop to as low as 8 beats per minute during the winter. What is the percent of change in a bear’s heart rate from the summer rate to the winter low rate? EXAMPLE 1 Finding a Percent of Decrease SOLUTION To find the percent of decrease, use the percent of change equation.
p% = = 52 _ = 60 0.86 86.67% 60 – 8 ANSWER 60 The percent of decrease is about 86.7%. EXAMPLE 1 Finding a Percent of Decrease Write amount of decrease and divide by original amount. Subtract. Write fraction as a decimal. Write decimal as a percent.
School Enrollment A school had 825 students enrolled last year. This year, 870 students are enrolled. Find the percent of increase to the nearest tenth. 870 – 825 p% = 825 45 = 825 __ 0.0545 = 5.45% EXAMPLE 2 Finding a Percent of Increase SOLUTION Write amount of increase and divide by original amount. Subtract. Write fraction as a decimal. Write decimal as a percent.
ANSWER The percent of increase is about 5.5%. EXAMPLE 2 Finding a Percent of Increase
National Parks From October to November one year, attendance at Everglades National Park increased about 27.4%. There were 59,084 visitors in October. To the nearest hundred, how many people visited in November? STEP 1 Find the increase. Increase 27.4%of 59,084 = = 0.274( 59,084) 16,189 EXAMPLE 3 Using Percent of Increase SOLUTION Write 27.4% as a decimal. Multiply.
Add the increase to the original amount. STEP 2 New Amount 16,189 75,273 59,084 = + EXAMPLE 3 Using Percent of Increase ANSWER About 75,300people visited the park in November.
1. Original amt = 50 50 – 36 p% = 50 14 = 50 0.28 = 28% = for Examples 1, 2 and 3 GUIDED PRACTICE Tell whether the change is an increase or decrease. Then find the percent of change. Round to the nearest percent. New amt = 36 The change is a decrease since the new amount is less than the original amount Write amount of decrease and divide by original amount. Subtract. Write fraction as a decimal. Write decimal as a percent.
2. Original amt = 10 29.5 – 10 = 10 p% 19.5 = 10 1.95 = 195% = for Examples 1, 2 and 3 GUIDED PRACTICE New amt = 29.5 The change is an increase since the new amount is greater than the original amount Write amount of decrease and divide by original amount. Subtract. Write fraction as a decimal.
3. Original amt = 90 110 – 90 = 90 p% 20 = 90 0.222 = 22% = for Examples 1, 2 and 3 GUIDED PRACTICE New amt = 110 The change is an increase since the new amount is greater than the original amount Write amount of decrease and divide by original amount. Subtract. Write fraction as a decimal.
4. What If? In Example 3, suppose the attendance decreased by about 11%. Approximate, to the nearest thousand, the attendance in November decrease = 11% of 59,084 for Examples 1, 2 and 3 GUIDED PRACTICE SOLUTION Find the decrease. = 0.11(59,084) = 6499.24
New amount 59,084 – 6499 52,585 ANSWER About 53,000people visited the park in November. for Examples 1, 2 and 3 GUIDED PRACTICE Subtract the decrease to the original amount.