Chapter 3 Lesson 1

Chapter 3:

Solve each problem and express the answer in scientific notation

a. (6.6 ❌ 10^-8) + (5 ❌10^-9)

6.6E-8+5E-9=7.1E-8

                6.6❌10^-8+.5❌10^-8... 6.6+.5= 7.1 ❌10^-8

Answer: 7.1E-8

b. (9.4 ❌10^-2) - (2.1 ❌ 10^-2)

9.4E-2-2.1E-2= .073

                9.4-2.1=7.3❌10^-2 or .073

Answer: 7.3E-2

2. Calculate the following and write your answer in scientific notation: 

                 6.6❌10^6➗

(8.8❌10^-2) ❌ (2.5❌10^3)

6.6❌10^6=6.6E6

(8.8E-2) ❌ (2.5E3)= 220 or 2.2❌10^2

6.6❌10^6 ➗ 2.2❌10^2= 30000 or. 3E4

Answer from trying it by hand: 3E4

From calculador: 1.9E11

3. The student measures the depth of a swimming pool to be 2.04 meters at its deepest end. The accepted value is 2.0 meters. What's the students percent error. 2.04-2.0➗2.0❌100=.2. The percentage of error is .2%

13. Accuracy is a measure of how close a measurement comes to the actual or true value of whatever is measured. Precision is a measure of how close a series of measurements are to one another, irrespective of the actual value. You can measure accuracy by comparing it to the correct answer. Precision can be measured by comparing it to two or more other related measurements.

15. Calculate A technician experimentally determined the boiling point of octane to be 124.1ºC. The actual boiling point of octane is 125.7ºC. Calculate the error and the percent error. 124.1-125.7➗125.7❌100= 1.273 rounded. The percentage of error rounded is 1.273%

17. Calculate Solve the following and express each answer in scientific notation and to the correct number of significant figures.

a. (5.3 ❌ 10^4) + (1.3❌10^4) 

6.6❌10^4=66000

Answer+ 6.6E4

b. (7.2 ❌ 10^-4) ➗ (1.8❌10^3)

7.2➗1.8❌10^-7= 4E-7

Answer: 4E-7

c. 10^4 ❌ 10^−3 ❌ 10^6

10^4 ❌ 10^−3 ❌ 10^6= 1E10

Answer: 1E10

d. (9.12 ❌ 10^−1) -  (4.7 ❌ 10^−2)

(9.12E-1)-(4.7E-2)= .865

                9.12-.47=8.65^-1

Answer: 8.65E-1

e. (5.4❌ 10^4) ❌ (3.5 ❌ 10^9)

(5.4E4)❌(3.5E9)= 1.89E14

                18.9❌10^13=18.9 move decimal and round

Answer: 1.9E14