Review for the first midterm

...

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

Significant digits; rounding; absolute and relative errors; rounding after an operation.

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