Section 3B

Putting numbers in perspective ... exercises

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Review of Scientific Notations (p.136)

Exercise 15 p.148

Convert from scientific to ordinary notations

\(3\times 10^3\)=3,000

\(6\times 10^6\)=6,000,000

\(3.4\times 10^5\)=340,000

\(2\times 10^{-2}\)=0.02

\(2.1\times 10^{-4}\)=0.00021

\(4\times 10^{-5}\)=0.00004

Review of Scientific Notations (p.136)

Exercise 17 p.148

Convert from ordinary to scientific notations

233=\(2.33 \times 10^2 \)

126,547=\(1.26547 \times 10^5 \)

0.11=\(1.1 \times 10^{-1} \)

9736.23=\(9.73623 \times 10^{3} \)

124.58=\(1.2458 \times 10^2 \)

0.8642=\(8.642 \times 10^{-1} \)

Review of Scientific Notations (p.136)

Exercise 19 p.148

Perform the operations; express the answers in scientific notation

\( (3 \times 10^3) \times (2 \times 10^2) \) \(= (3 \times 2) \times (10^3 \times 10^2) \) \( = 6 \times 10^5 \) <------ your answer

Exercise 19 p.148 ... continued

\( (4 \times 10^2) \times (3 \times 10^8) \) \(= (4 \times 3) \times (10^2 \times 10^8) \) \( = 12 \times 10^{10} \) \( = 1.2 \times 10^{11} \) <------ your answer

\( (3 \times 10^3) + (2 \times 10^2) \) \(= (3 \times 10^3) + (0.2 \times 10^3) \) \(= (3+0.2) \times 10^3 \) \(= 3.2 \times 10^3 \) <------ your answer

\( (8 \times 10^{12}) \div (2 \times 10^4) \) \( = \frac{8 \times 10^{12}} {2 \times 10^4} \) \(= \frac{8}{2} \times \frac{10^{12}}{10^4} \)
\(= 4 \times 10^8\) <------ your answer

Review of Scientific Notations (p.136)

Exercise 21 p.148

Give the factor by which the numbers differ

\(10^{35} \) and \(10^{26}\)

The factor is their ratio: \(\frac{10^{35}}{10^{26}} = 10^{35-26}=10^9 \)

Answer: \(10^{35}\) is \(10^9\) times \(10^{26} \).

Exercise 21 p.148 ... continued

Give the factor by which the numbers differ

\(10^{17} \) and \(10^{27}\)

The factor is their ratio: \(\frac{10^{17}}{10^{27}} = 10^{17-27}=10^{-10} \)

Answer: \(10^{17}\) is \(10^{27}\) times \(10^{-10} \).

Exercise 21 p.148 ... continued

Give the factor by which the numbers differ

1 billion and 1 million

1 billion is \(10^9\) while 1 million is \(10^6\).

The factor is their ratio: \(\frac{10^{9}}{10^{6}} = 10^{9-6}=10^{3} \)

Answer: 1 billion is a thousand times bigger than 1 million.

Using Scientific Notations

Exercise 23 p.148

Rewrite the statement using scientific notations.

My new music player has a capacity of 340 gigabytes.

Since "giga" means billion, 340 gigabytes is 340 billion bytes,

that is \(340 \times 10^9\) \(= 3.4 \times 10^{11} \)

We conclude that the music player has a capacity of \(3.4 \times 10^{11} \) bytes.

Using Scientific Notations

Exercise 25 p.148

Rewrite the statement using scientific notations.

The diameter of a typical bacterium is about 0.000001 meter.

Since \( 0.000001 = 10^{-6} \) this diameter is about

\(10^{-6}\) meter, or one micrometer.

Approximation with Scientific Notations

Exercise 27 p.148

Estimate the quantity \(300,000 \times 100 \).

\(300,000 \times 100 \) \(=3 \times 10^5 \times 10^2\) \(=3 \times 10^7\) <---- your answer

This value is exact.

Exercise 27 p.148 ... continued

Estimate the quantitiy 5.1 million \(\times\) 1.9 thousand.

Note that 5.1 million \(=5.1 \times 10^6 \),
while 1.9 thousand \(=1.9 \times 10^3 \)

We now multiply \( (5.1 \times 10^6) \times (1.9 \times 10^3)\) \(=(5.1 \times 1.9) \times 10^9\) \(\approx 10 \times 10^9\) \(=10^{10}\) <------- your approximation

The exact value is \(9.69 \times 10^9 \).

Exercise 27 p.148 ... continued

Estimate the quantitiy \(4 \times 10^9 \div (2.1 \times 10^6)\)

We divide \(\frac{4 \times 10^9} {2.1 \times 10^6} \) \(=\frac{4}{2.1} \times \frac{10^9}{10^6}\) \( \approx 2 \times 10^3 \) <------- your approximation

The exact value is \(1,904.762 \approx 1.9 \times 10^3 \).

Order of magnitude estimate.

Exercise 33 p.148

Estimate the number of times your heart beats in a week.

Solution. One has approximately one heartbeat in a second. There are 60 seconds in a minute, 60 minutes in an hour, and 24 hours in a day, and 7 days in a week. We thus need to approximately calculate \(60 \times 60 \times 24 \times 7 \) \(= 604800 \) \( \approx 6 \times 10^5 \) <----- your answer

Scale ratios.

Exercise 49 p.149

Find the scale ratio on a map where 2 centimeters represent 100 kilometers.

Since centi- means \(10^{-2}\), 2 centimeters is \(2 \times 10^{-2} \) meters.

Since kilo- means \(10^3\), 100 kilometers is \(100 \times 10^3\) \(= 10^5 \) meters.

Their ratio: \( \frac{10^5}{2 \times 10^{-2}} \) \(= \frac{1 \times 10^5}{2 \times 10^{-2}} \) \(= 0.5 \times 10^7\) \(= 5 \times 10^6 \) \(= 5,000,000 \)
Answer: Scale ratio is \( 5,000,000 \) to \(1 \)

Exercise 50 p.149

Find the scale ratio on a map where 1 inch represent 10 miles.

We need to calculate the ratio \(\frac{10 \ \text{mile}}{1 \ \text{inch}}\).

1 mile = 1,760 yards, 1 yard = 3 feet, 1 foot = 12 inches.

Thus 10 mile = \( 10 \times 1,760 \times 3 \times 12 = 633,600 \) inches.

Answer: Scale ratio is \( 633,600 \) to \(1 \)

Making numbers understandable

Exercise 59 p.149

Approximately 32,300 Americans died in automobile accidents in 2012. Express this toll in deaths per day.

There are 365 days in a year.

The average number of deaths per day is \( \frac {32300}{365} \approx 88 \) <------ your answer

Making numbers understandable

Exercise 70 p.150

Bad news: the Sun is going to die in approximately 5 billion years.

It took approximately 65 million years from the time the dinosaurs were wiped out by an asteroid impact until humans arrived on the scene.

Assume that by an abuse of technology humans are wiped out, and it takes another 65 million years for the next intelligent species to arise on Earth. They again abuse technology to wipe themselves out.

How many such cycles may occur until the Sun dies?

Exercise 70 p.150 ... continued

65 million years from a catastrophe till intelligent species, and 5 billion years until the Sun dies

Solution: \( \frac {5 \ \text {billion}}{65 \ \text{million}} = \frac{5 \times 10^9}{65 \times 10^6} = \frac{5}{65} \times 10^3 \approx 77 \)

Answer: approximately 77 times