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## Section 3A

### Percentages:basic skills and concepts; solving percentage problems

##### Percentages:basic skills and concepts

Exercise 30 p.132

Express the percentage $$121 \%$$ as a reduced fraction

Solution

Divide it by $$100$$ to get

$$121\% = \frac{121}{100}$$

Answer: $$121\%$$ is equivalent to $$\frac{121}{100}$$

##### Percentages:basic skills and concepts

Exercise 30 p.132

Express the percentage $$121 \%$$ as a decimal

Solution

Divide it by $$100$$ to get

$$121\% = \frac{121}{100} =1.21$$

Answer: $$121\%$$ is equivalent to $$1.21$$

##### Percentages:basic skills and concepts

Exercise 51 p.133

The average sale price of a house in the US decreased from $$301,000$$ in February $$2008$$ to $$\ 152,000$$ in February $2013$. Find the percentage change.

Solution

The absolute change is $$152000 - 301000 = - 149000$$

Divide the absolute change by the reference value of $$\301,000$$:

$$\text{ relative change }= \frac{\text{absolute change}}{\text{reference value}}$$ $$= \frac{-149000}{301000} = -0.495$$

Express this value as percentage: $$-0.495 = -49.5\%$$

##### Percentages:basic skills and concepts

Exercise 51 p.133 ... variation

The average sale price of a house in the US decreased from $$301,000$$ in February $$2008$$ to $$\ 152,000$$ in February $2013$. By which percentage was an average house more expansive in $2008$ than in $2013$?

Solution

While the absolute change is still $$\ 149000$$, the reference value is now $$\152000$$:

$$\text{ relative change }$$ $$= \frac{149000}{152000} = 0.980 =98.0 \%$$

While the price has dropped by $$49.5 \%$$, the house was more expansive by $$98.0\%$$ < ----------- shift of reference value

##### Percentages:solving percentage problems

Exercise 76 p.133

The final cost of your new shoes is $$\ 107.69$$. The local sales tax rate is $$6.2 \%$$. What was the retail (pre-tax) price?

Solution

The price of $$\ 107.69$$ constitutes $$106.2 \% =1.062$$ of the quantity in question.

The quantity in question thus is $$107.69 / 1.062 = \ 101.40$$

## Section 3B

### Scientific notation; operations with numbers in scientific notation; scale ratios.

##### Scientific notation: recall from Section 3B

$$Q \times \text{a power of 10}$$,

$$Q$$ --- a quantity between 1 and 10 ,

a power of 10 is $$10 ^ \text{a whole number}$$

Some powers of ten have their own names:

 $$10^2$$ | hundred $$10^3$$ | thousand $$10^6$$ | million $$10^9$$ | billion $$10^{12}$$ | trillion
##### Exercise 18 p.148

Write each of the following numbers in scientific notation:

$$4327$$ $$= 4.327 \times 10^3$$

$$984.35$$ $$= 9.8435 \times 10^2$$

$$0.0045$$ $$= 4.5\times 10^{-3}$$

$$624.87$$ $$= 6.2487 \times 10^2$$

$$0.1357$$ $$= 1.357 \times 10^{-1}$$

$$98.180004$$ $$= 9.81880004 \times 10$$

##### Exercise 20 p.148

Do the calculation and express the answer in scientific notation

$$(4 \times 10^7) \times (2 \times 10^8)$$ $$= (4\times 2) \times (10^7 \times 10^8)$$
$$= 8$$ $$\times 10^{15}$$

$$(3.2 \times 10^5) \times (2 \times 10^4)$$ $$= (3.2\times 2) \times (10^5 \times 10^4)$$
$$= 6.4$$ $$\times 10^{9}$$

$$(4 \times 10^3) + (5 \times 10^2)$$ $$= (40 \times 10^2) + (5 \times 10^2)$$
$$= (40+5)$$ $$\times 10^{2}$$ $$= (45)$$ $$\times 10^{2}$$ $$= 4.5$$ $$\times 10^{3}$$

$$(9 \times 10^{13}) \div (3 \times 10^{10})$$ $$= \frac{ 9 \times 10^{13}}{ 3 \times 10^{10} }$$ $$= 3$$ $$\times 10^{3}$$

##### Exercise 49 p.149 ... modified

Find the scale ratio for the map where 4 centimeters on the map represent 50 kilometers.

Solution.

Since centi- means $$10^{-2}$$, 4 centimeters is $$4 \times 10^{-2}$$ meters.

Since kilo- means $$10^3$$, 50 kilometers is $$50 \times 10^3$$ $$= 5 \times 10^4$$ meters.

Their ratio: $$\frac{5 \times 10^4}{4 \times 10^{-2}}$$ $$= \frac{500 \times 10^2}{4 \times 10^{-2}}$$ $$= 125 \times 10^4$$ $$= 1.25 \times 10^6$$ $$= 1,250,000$$
Answer: Scale ratio is $$1,250,000$$ to $$1$$

## Section 3C

### Counting Significant Digits

Exercises 17-28 p.160 ... modified ... State the number of significant digits

37 .... 2 significant digits

3.7 .... 2 significant digits

0.37 .... 2 significant digits

0.037 .... 2 significant digit

0.0370 .... 3 significant digits

2.037 .... 4 significant digits

##### Exercise 49 p.160

Your speedometer reads 60 miles per hour when you are actually traveling 58 miles per hour. Find the absolute and relative errors.

$$\text{absolute error} = \text{measured value} - \text{true value}$$
$$= 60 - 58$$ $$= 2$$ mi/hr <--- your answer (absolute error)

$$\text{relative error} = \frac{\text{absolute error}}{\text{true value}} \times 100 \%$$ $$= \frac{2}{58} \times 100 \%$$ $$= 0.03448 \times 100$$ $$= 3.4\%$$ <--- your answer (relative error)

##### Exercise 60 p.161 ... modified

How far will you travel driving at a speed of $43$ miles per hour during $0.25$ hours? Use appropriate rounding rules to express the result with a correct number of significant digits.

Solution.

We find
$$43 \times 0.25 = 10.75$$.

The number $10.75$ has for significant digits while we must leave only $2$.

We round it: $$10.75 \approx 11$$.

Answer: $11$ miles.

## Section 3E

### Better in each case but worse overall

##### Better drug. Exercise 22 p.181 ... modified
Two drugs, A and B, were tested.
Women Men
Drug A 10 of 200 cured 800 of 1600 cured
Drug B 108 of 1080 cured 432 of 720 cured

Drug A administered to: 200+1600=1800.

Cured by drug A: 10+800=810 (out of 1800).

Drug B administered to: 1080+720=1800.

Cured by drug B: 108+432=540 (out of 1800).

##### Cured by the drugs:
Women Men Total
Drug A 10 of 200 800 of 1600 810 of 1800
Drug B 108 of 1080 432 of 720 540 of 1800
##### Women cured:
Women Men Total
Drug A 10 of 200 800 of 1600 810 of 1800
Drug B 108 of 1080 432 of 720 540 of 1800

Drug A: $$10 \div 200 = \frac{10}{200}$$ $$=\frac{1}{20} = 5\%$$

Drug B: $$108 \div 1080 = \frac{108}{1080}$$ $$=\frac{1}{10} = 10\%$$

##### Women cured:
Women Men Total
Drug A 5% 800 of 1600 810 of 1800
Drug B 10% 432 of 720 540 of 1800
##### Men cured:
Women Men Total
Drug A 5% 800 of 1600 810 of 1800
Drug B 10% 432 of 720 540 of 1800

Drug A: $$800 \div 1600 = \frac{800}{1600}$$ $$=\frac{1}{2} = 50\%$$

Drug B: $$432 \div 720 = \frac{432}{720}$$ $$=\frac{6}{10} = 60\%$$

##### Men cured:
Women Men Total
Drug A 5% 50% 810 of 1800
Drug B 10% 60% 540 of 1800
##### Total cured:
Women Men Total
Drug A 5% 50% 810 of 1800
Drug B 10% 60% 540 of 1800

Drug A: $$810 \div 1800$$ $$= 0.45 = 45\%$$

Drug B: $$540 \div 1800$$ $$=0.3=30\%$$

##### Total cured:
Women Men Total
Drug A 5% 50% 45%
Drug B 10% 60% 30%
##### Compare two tables
The initial one was:
Women Men
Drug A 10 of 200 cured 800 of 1600 cured
Drug B 108 of 1080 cured 432 of 720 cured
We obtained:
Women Men Total
Drug A 5% 50% 45%
Drug B 10% 60% 30%

In the study, drug B was administered to an unproportionally big amount of women; that corrupted the overall results