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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 \( \frac{121}{100} \)
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 \( 1.21 \)
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: \(-0.495 = -49.5\%\)
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
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 \( 107.69 / 1.062 = \$ 101.40 \)
\( Q \times \text{a power of 10} \),
\( Q \) --- 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 |
Write each of the following numbers in scientific notation:
\(4327 \) \( = 4.327 \times 10^3 \)
\(984.35 \) \( = 9.8435 \times 10^2 \)
\(0.0045 \) \( = 4.5\times 10^{-3} \)
\(624.87 \) \( = 6.2487 \times 10^2 \)
\(0.1357 \) \( = 1.357 \times 10^{-1} \)
\(98.180004 \) \( = 9.81880004 \times 10 \)
Do the calculation and express the answer in scientific notation
\( (4 \times 10^7) \times (2 \times 10^8)\)
\( = (4\times 2) \times (10^7 \times 10^8) \)
\( = 8 \)
\( \times 10^{15} \)
\( (3.2 \times 10^5) \times (2 \times 10^4)\)
\( = (3.2\times 2) \times (10^5 \times 10^4) \)
\( = 6.4 \)
\( \times 10^{9} \)
\( (4 \times 10^3) + (5 \times 10^2)\)
\( = (40 \times 10^2) + (5 \times 10^2) \)
\( = (40+5) \)
\( \times 10^{2} \)
\( = (45) \)
\( \times 10^{2} \)
\( = 4.5 \)
\( \times 10^{3} \)
\( (9 \times 10^{13}) \div (3 \times 10^{10})\) \( = \frac{ 9 \times 10^{13}}{ 3 \times 10^{10} }\) \( = 3 \) \( \times 10^{3} \)
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 \(1 \)
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
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)
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.
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).
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 | 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 | Men | Total | |
---|---|---|---|
Drug A | 5% | 800 of 1600 | 810 of 1800 |
Drug B | 10% | 432 of 720 | 540 of 1800 |
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: \(432 \div 720 = \frac{432}{720} \) \(=\frac{6}{10} = 60\%\)
Women | Men | Total | |
---|---|---|---|
Drug A | 5% | 50% | 810 of 1800 |
Drug B | 10% | 60% | 540 of 1800 |
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: \(540 \div 1800 \) \(=0.3=30\%\)
Women | Men | Total | |
---|---|---|---|
Drug A | 5% | 50% | 45% |
Drug B | 10% | 60% | 30% |
Women | Men | |
---|---|---|
Drug A | 10 of 200 cured | 800 of 1600 cured |
Drug B | 108 of 1080 cured | 432 of 720 cured |
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