Number & Algebra

Using multiplication in scientific notation, total mass $=$ (mass per atom) $\times$ (number of atoms): $$\text{Total mass} = (1.67 \times 10^{-27}) \times (3.5 \times 10^{24})$$ Multiply the coefficients and add the exponents: $$= (1.67 \times 3.5) \times 10^{-27+24}$$ $$= 5.845 \times 10^{-3} \text{ kg}$$ Since $5.845$ already lies between 1 and 10, the form is already standard. Rounding the coefficient to 3 significant figures: $$\boxed{\text{Total mass} \approx 5.85 \times 10^{-3} \text{ kg}}$$

Using profit $=$ revenue $-$ costs, first rewrite both quantities with the same power of 10. $$\text{Revenue} = 4.2 \times 10^{7} = 42 \times 10^{6}$$ $$\text{Costs} = 9.8 \times 10^{6}$$ Substituting into the subtraction: $$\text{Profit} = 42 \times 10^{6} - 9.8 \times 10^{6}$$ $$= (42 - 9.8) \times 10^{6} = 32.2 \times 10^{6}$$ The coefficient $32.2$ is not between 1 and 10, so convert to correct scientific notation: $$32.2 \times 10^{6} = 3.22 \times 10^{7}$$ $$\boxed{\text{Profit} = 3.22 \times 10^{7} \text{ USD}}$$ (This value is already exact to 3 significant figures, so no further rounding is needed.)

(a) Using time $=$ (number of years) $\times$ (seconds per year): $$\text{Time} = 4.25 \times (3.156 \times 10^{7})$$ $$= (4.25 \times 3.156) \times 10^{7}$$ Multiplying the coefficients: $$4.25 \times 3.156 = 13.413$$ $$\text{Time} = 13.413 \times 10^{7}$$ Converting to standard form: $$\text{Time} = 1.3413 \times 10^{8} \text{ s}$$ $$\boxed{\text{Time} \approx 1.34 \times 10^{8} \text{ s (3 s.f.)}}$$ (b) Using distance $=$ speed $\times$ time, and the unrounded value from (a) for accuracy: $$\text{Distance} = (3.00 \times 10^{8}) \times (1.3413 \times 10^{8})$$ Multiply the coefficients and add the exponents: $$3.00 \times 1.3413 = 4.0239, \qquad 8+8=16$$ $$\text{Distance} = 4.0239 \times 10^{16} \text{ m}$$ Converting metres to kilometres (dividing by $10^{3}$): $$\text{Distance} = 4.0239 \times 10^{13} \text{ km}$$ $$\boxed{\text{Distance} \approx 4.02 \times 10^{13} \text{ km (3 s.f.)}}$$

(a) Using time $=$ total operations $\div$ operations per second: $$\text{Time} = \frac{5.4 \times 10^{13}}{3.6 \times 10^{9}}$$ Divide the coefficients and subtract the exponents: $$5.4 \div 3.6 = 1.5, \qquad 13-9=4$$ $$\text{Time} = 1.5 \times 10^{4} \text{ s}$$ $$\boxed{\text{Time} = 1.5 \times 10^{4} \text{ s (exact)}}$$ (b) Using the conversion 1 hour $= 3600$ s: $$\text{Time (hours)} = \frac{1.5 \times 10^{4}}{3600} = \frac{15\,000}{3600}$$ $$= 4.1\overline{6}$$ $$\boxed{\text{Time} \approx 4.17 \text{ hours (3 s.f.)}}$$

(a) Using the volume formula $V = \dfrac{4}{3}\pi r^{3}$, first find the radius from the diameter. $$r = \frac{1.2 \times 10^{-7}}{2} = 6 \times 10^{-8} \text{ m}$$ Cubing the radius: $$r^{3} = (6 \times 10^{-8})^{3} = 6^{3} \times 10^{-24} = 216 \times 10^{-24} = 2.16 \times 10^{-22}$$ Substituting into the volume formula: $$V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi \times 2.16 \times 10^{-22}$$ Using $\dfrac{4}{3}\pi = 4.18879\ldots$: $$V = 4.18879 \times 2.16 \times 10^{-22} = 9.0478 \times 10^{-22} \text{ m}^3$$ $$\boxed{V \approx 9.05 \times 10^{-22} \text{ m}^3 \text{ (3 s.f.), as required}}$$ (b) Using number of particles $=$ (total volume) $\div$ (volume of one particle), first convert $1\text{ cm}^3$ to $\text{m}^3$, then divide by the unrounded value from (a) for accuracy. $$1 \text{ cm}^{3} = 1 \times 10^{-6} \text{ m}^{3}$$ $$\text{Number of particles} = \frac{1 \times 10^{-6}}{9.0478 \times 10^{-22}}$$ Divide coefficients and subtract exponents: $$= \frac{1}{9.0478} \times 10^{-6-(-22)} = 0.110527 \times 10^{16}$$ Converting to standard form: $$= 1.10527 \times 10^{15}$$ $$\boxed{\text{Number of particles} \approx 1.11 \times 10^{15} \text{ (3 s.f.)}}$$

Using logarithms to solve for the exponent, take $\log$ (or $\ln$) of both sides and apply the power law of logarithms to bring down the exponent. $$5^{x} = 17$$ $$\ln(5^{x}) = \ln(17)$$ By the power law, $\ln(5^x) = x\ln 5$: $$x\ln(5) = \ln(17)$$ Solving for $x$: $$x = \frac{\ln 17}{\ln 5} = \frac{2.833213}{1.609438}$$ $$x = 1.76037\ldots$$ $$\boxed{x \approx 1.76 \text{ (3 s.f.)}}$$

Using the subtraction law of logarithms, $\log_a(m) - \log_a(n) = \log_a\left(\dfrac{m}{n}\right)$, to combine the left-hand side into a single logarithm. $$\log_3(x) - \log_3(4) = 2$$ $$\log_3\left(\frac{x}{4}\right) = 2$$ Converting to exponential form: $$\frac{x}{4} = 3^{2} = 9$$ Solving for $x$: $$x = 4 \times 9$$ $$\boxed{x = 36}$$

Using logarithms: take $\ln$ of both sides, apply the power law to bring down each exponent, then solve the resulting linear equation for $x$. $$2^{x+1} = 3^{x-1}$$ $$\ln(2^{x+1}) = \ln(3^{x-1})$$ $$(x+1)\ln 2 = (x-1)\ln 3$$ Substituting $\ln 2 = 0.693147$ and $\ln 3 = 1.098612$: $$0.693147(x+1) = 1.098612(x-1)$$ Expanding both sides: $$0.693147x + 0.693147 = 1.098612x - 1.098612$$ Collecting $x$ terms on one side and constants on the other: $$0.693147 + 1.098612 = 1.098612x - 0.693147x$$ $$1.791759 = 0.405465x$$ Solving for $x$: $$x = \frac{1.791759}{0.405465} = 4.41903\ldots$$ $$\boxed{x \approx 4.42 \text{ (3 s.f.)}}$$

(a) Substituting directly into the given formula with $I = 1\times10^{2}$ and $I_0 = 1\times10^{-12}$: $$L = 10\log_{10}\left(\frac{1\times10^{2}}{1\times10^{-12}}\right) = 10\log_{10}(1\times10^{14})$$ Since $\log_{10}(1\times10^{14}) = 14$: $$L = 10 \times 14$$ $$\boxed{L = 140 \text{ dB}}$$ (b) Substituting $L=85$ into the formula and rearranging to make $I$ the subject, converting from logarithmic to exponential form: $$85 = 10\log_{10}\left(\frac{I}{1\times10^{-12}}\right)$$ Dividing both sides by 10: $$\log_{10}\left(\frac{I}{1\times10^{-12}}\right) = 8.5$$ Converting to exponential form: $$\frac{I}{1\times10^{-12}} = 10^{8.5}$$ $$I = 10^{8.5} \times 10^{-12} = 10^{-3.5}$$ Evaluating: $$10^{-3.5} = 3.16228\times10^{-4}$$ $$\boxed{I \approx 3.16\times10^{-4} \text{ W/m}^2 \text{ (3 s.f.)}}$$

(a) Substituting both data points into the model $N(t) = A \cdot b^t$ to form two equations, then dividing one by the other to eliminate $A$: $$N(2) = Ab^{2} = 5000$$ $$N(6) = Ab^{6} = 12\,800$$ Dividing the second equation by the first: $$\frac{Ab^{6}}{Ab^{2}} = \frac{12\,800}{5000}$$ $$\boxed{b^{4} = 2.56 \text{, as required}}$$ Taking the positive fourth root (since $b>0$ for a realistic growth model): $$b = 2.56^{1/4} = \sqrt{\sqrt{2.56}} = \sqrt{1.6}$$ $$\boxed{b = \sqrt{1.6} \approx 1.26 \text{ (3 s.f.)}}$$ (b) Substituting $b^{2} = \sqrt{2.56} = 1.6$ back into $Ab^{2}=5000$: $$A(1.6) = 5000$$ Solving for $A$: $$A = \frac{5000}{1.6}$$ $$\boxed{A = 3125}$$ (c) Substituting $t=10$ into the model, using the exact form $b^2 = 1.6$ for accuracy: $$N(10) = A \cdot b^{10} = 3125 \times (b^{2})^{5} = 3125 \times (1.6)^{5}$$ Evaluating the power: $$1.6^{5} = 10.48576$$ $$N(10) = 3125 \times 10.48576$$ $$N(10) = 32\,768$$ $$\boxed{N(10) \approx 32\,768 \text{ active users}}$$

(a) The length is given to the nearest 0.1 m, so the true length lies within $0.05$ m of the stated value. $$\text{Lower bound} = 3.6 - 0.05 = 3.55 \text{ m}$$ $$\text{Upper bound} = 3.6 + 0.05 = 3.65 \text{ m}$$ $$\boxed{3.55 \text{ m} \leq \text{length} < 3.65 \text{ m}}$$ (b) For 5 planks placed end to end, multiply each bound by 5. $$\text{Lower bound (total)} = 5 \times 3.55 = 17.75 \text{ m}$$ $$\text{Upper bound (total)} = 5 \times 3.65 = 18.25 \text{ m}$$ $$\boxed{17.75 \text{ m} \leq \text{total length} < 18.25 \text{ m}}$$

(a) Using absolute error $= |\text{approximate value} - \text{exact value}|$: $$\text{Absolute error} = |9.8 - 9.80665|$$ $$\boxed{\text{Absolute error} = 0.00665 \text{ m s}^{-2}}$$ (b) Using percentage error $= \dfrac{\text{absolute error}}{\text{exact value}} \times 100\%$: $$\text{Percentage error} = \frac{0.00665}{9.80665} \times 100$$ Evaluating using a GDC: $$\frac{0.00665}{9.80665} = 0.0006782\ldots$$ $$= 0.06782\ldots\%$$ $$\boxed{\text{Percentage error} \approx 0.0678\% \text{ (3 s.f.)}}$$

(a) Each measurement is given to the nearest 0.1 m, so each true value lies within 0.05 m of the stated value. $$\text{Length: lower bound} = 8.35 \text{ m}, \quad \text{upper bound} = 8.45 \text{ m}$$ $$\text{Width: lower bound} = 5.15 \text{ m}, \quad \text{upper bound} = 5.25 \text{ m}$$ (b) The minimum possible area occurs at the lower bounds of both dimensions, and the maximum possible area occurs at the upper bounds of both dimensions. $$\text{Lower bound area} = 8.35 \times 5.15 = 43.0025 \text{ m}^2$$ $$\text{Upper bound area} = 8.45 \times 5.25 = 44.3625 \text{ m}^2$$ $$\boxed{43.0025 \text{ m}^2 \leq \text{area} \leq 44.3625 \text{ m}^2}$$ (c) The calculated (nominal) area uses the rounded measurements directly: $$\text{Calculated area} = 8.4 \times 5.2 = 43.68 \text{ m}^2$$ The largest possible deviation from this calculated value occurs at whichever bound of the true area is furthest away; comparing the two differences: $$44.3625 - 43.68 = 0.6825, \qquad 43.68 - 43.0025 = 0.6775$$ Since $0.6825 > 0.6775$, the maximum absolute error is: $$\text{Maximum absolute error} = 44.3625 - 43.68 = 0.6825 \text{ m}^2$$ Using percentage error $= \dfrac{\text{maximum absolute error}}{\text{calculated area}} \times 100\%$: $$\text{Maximum percentage error} = \frac{0.6825}{43.68} \times 100 = 1.5625\ldots\%$$ $$\boxed{\text{Maximum percentage error} \approx 1.56\% \text{ (3 s.f.)}}$$

(a) Distance is given to the nearest 5 km, so the true distance lies within 2.5 km of the stated value; time is given to the nearest 0.1 hours, so the true time lies within 0.05 hours of the stated value. $$\text{Lower bound (distance)} = 150 - 2.5 = 147.5 \text{ km}, \quad \text{Upper bound (distance)} = 150 + 2.5 = 152.5 \text{ km}$$ $$\text{Lower bound (time)} = 2.5 - 0.05 = 2.45 \text{ hours}, \quad \text{Upper bound (time)} = 2.5 + 0.05 = 2.55 \text{ hours}$$ (b) Using average speed $=$ distance $\div$ time: the maximum speed occurs at maximum distance and minimum time, and the minimum speed occurs at minimum distance and maximum time. $$\text{Max speed} = \frac{152.5}{2.45} = 62.2448\ldots \approx 62.2 \text{ km h}^{-1} \text{ (3 s.f.)}$$ $$\text{Min speed} = \frac{147.5}{2.55} = 57.8431\ldots \approx 57.8 \text{ km h}^{-1} \text{ (3 s.f.)}$$ $$\boxed{57.8 \text{ km h}^{-1} \leq \text{speed} \leq 62.2 \text{ km h}^{-1} \text{ (3 s.f.)}}$$ (c) The calculated speed (using the rounded values given) is: $$\text{Calculated speed} = \frac{150}{2.5} = 60 \text{ km h}^{-1}$$ Comparing this to the maximum possible speed found in part (b), $62.2448\ldots \text{ km h}^{-1}$: $$\text{Percentage error} = \left|\frac{60 - 62.2448}{62.2448}\right| \times 100 = 3.606\ldots\%$$ $$\boxed{\text{Percentage error} \approx 3.74\% \text{ (3 s.f.)}}$$

(a) $m = 45.0$ g is given to 3 s.f. (precision to the nearest 0.1 g), and $V = 5.8 \text{ cm}^3$ is given to 2 s.f. (precision to the nearest 0.1 cm$^3$), so each true value lies within half of that precision of the stated value. $$\text{Lower bound } (m) = 44.95 \text{ g}, \quad \text{Upper bound } (m) = 45.05 \text{ g}$$ $$\text{Lower bound } (V) = 5.75 \text{ cm}^3, \quad \text{Upper bound } (V) = 5.85 \text{ cm}^3$$ (b) Using $\rho = \dfrac{m}{V}$: since $\rho$ increases with $m$ and decreases with $V$, the maximum density occurs at maximum $m$ and minimum $V$, and the minimum density occurs at minimum $m$ and maximum $V$. $$\text{Upper bound (density)} = \frac{45.05}{5.75} = 7.8348\ldots \approx 7.83 \text{ g cm}^{-3} \text{ (3 s.f.)}$$ $$\text{Lower bound (density)} = \frac{44.95}{5.85} = 7.6838\ldots \approx 7.68 \text{ g cm}^{-3} \text{ (3 s.f.)}$$ $$\boxed{7.68 \text{ g cm}^{-3} \leq \rho \leq 7.83 \text{ g cm}^{-3} \text{ (3 s.f.)}}$$ (c) The reported (calculated) density, using the given rounded values directly, is: $$\text{Reported density} = \frac{45.0}{5.8} = 7.75862\ldots \text{ g cm}^{-3}$$ The maximum possible deviation from this value is given by the upper bound of the true density found in (b): $$\text{Maximum absolute error} = 7.8348 - 7.75862 = 0.07618$$ Using percentage error $= \dfrac{\text{maximum absolute error}}{\text{reported density}} \times 100\%$: $$\text{Maximum percentage error} = \frac{0.07618}{7.75862} \times 100 = 0.9816\ldots\%$$ $$\boxed{\text{Maximum percentage error} \approx 0.982\% \text{ (3 s.f.)}}$$

Using the formula for the $n$th term of an arithmetic sequence, $u_n = u_1 + (n-1)d$, with $u_1 = 7$ and $d = 4$: $$u_{15} = u_1 + (15-1)d$$ Substituting the values: $$u_{15} = 7 + (14)(4)$$ $$u_{15} = 7 + 56$$ $$\boxed{u_{15} = 63}$$

Using the formula for the $n$th term of a geometric sequence, $u_n = u_1 r^{n-1}$, first identify the common ratio from the given terms. $$r = \frac{15}{5} = 3$$ Substituting $u_1 = 5$, $r = 3$, $n = 8$: $$u_8 = u_1 r^{8-1} = 5 \times 3^{7}$$ Evaluating the power: $$3^{7} = 2187$$ $$u_8 = 5 \times 2187$$ $$\boxed{u_8 = 10\,935}$$

Since the number of seats increases by a constant amount per row, the row sizes form an arithmetic sequence with $u_1 = 18$, $d = 3$, $n = 20$. (a) Using the formula for the $n$th term, $u_n = u_1 + (n-1)d$: $$u_{20} = 18 + (20-1)(3)$$ $$u_{20} = 18 + (19)(3) = 18 + 57$$ $$\boxed{u_{20} = 75 \text{ seats}}$$ (b) Using the arithmetic series formula $S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right)$: $$S_{20} = \frac{20}{2}\left(2(18) + (19)(3)\right)$$ Simplifying inside the brackets: $$S_{20} = 10\left(36 + 57\right) = 10(93)$$ $$\boxed{S_{20} = 930 \text{ seats}}$$

The population after $n$ complete hours forms a geometric sequence, $P_n = 500 \times 1.08^{n}$. Using this to set up the inequality $P_n > 2000$: $$500 \times 1.08^{n} > 2000$$ Dividing both sides by 500: $$1.08^{n} > \frac{2000}{500} = 4$$ Taking $\ln$ of both sides (or using a GDC table/solver): $$n\ln(1.08) > \ln(4)$$ $$n > \frac{\ln 4}{\ln 1.08} = \frac{1.38629}{0.076961} = 18.013\ldots$$ Since $n$ must be a whole number of hours, check $n=18$ and $n=19$ using a GDC: $$n=18: \quad P_{18} = 500 \times 1.08^{18} = 500 \times 3.99557 = 1997.79 \text{ (not yet exceeded 2000)}$$ $$n=19: \quad P_{19} = 500 \times 1.08^{19} = 500 \times 4.31522 = 2157.61 \text{ (exceeds 2000)}$$ $$\boxed{n = 19 \text{ complete hours}}$$

Plan A is arithmetic with $u_1 = 40\,000$, $d = 1800$. Plan B is geometric with $u_1 = 38\,000$, $r = 1.05$. (a) Using $u_n = u_1 + (n-1)d$ for Plan A and $u_n = u_1 r^{n-1}$ for Plan B, with $n = 10$: $$\text{Plan A: } u_{10} = 40\,000 + (9)(1800) = 40\,000 + 16\,200$$ $$u_{10} = \$56\,200$$ $$\text{Plan B: } u_{10} = 38\,000 \times 1.05^{9}$$ $$u_{10} = 38\,000 \times 1.551328 = \$58\,950.47 \approx \$58\,950$$ (b) Using the arithmetic series formula $S_n = \dfrac{n}{2}\left(2u_1+(n-1)d\right)$ for Plan A, and the geometric series formula $S_n = \dfrac{u_1(r^n-1)}{r-1}$ for Plan B, both with $n=10$: $$\text{Plan A: } S_{10} = \frac{10}{2}\left(2(40\,000)+(9)(1800)\right)$$ $$= 5(80\,000+16\,200) = 5(96\,200) = \$481\,000$$ $$\text{Plan B: } S_{10} = \frac{38\,000(1.05^{10}-1)}{1.05-1}$$ Substituting $1.05^{10} = 1.628895$: $$S_{10} = \frac{38\,000(1.628895-1)}{0.05} = \frac{38\,000(0.628895)}{0.05}$$ $$= 38\,000 \times 12.57789 = \$477\,959.92 \approx \$477\,960$$ (c) Comparing the two totals: even though Plan B pays the higher salary by year 10 ($\$58\,950$ vs $\$56\,200$), Plan A produces the greater total earnings over the full 10 years ($\$481\,000$ vs $\$477\,960$), because Plan A's larger early salaries accumulate more before Plan B's growth compounds significantly. $$\boxed{\text{Plan A gives the greater total earnings, by approximately } \$3040 \text{ over 10 years.}}$$

Using the compound interest formula $FV = PV\left(1+\dfrac{r}{100}\right)^{n}$ (or equivalently the GDC finance/TVM solver with $N=6$, $PV=5000$, $I\% = 3.5$, $P/Y = C/Y = 1$): $$FV = 5000(1.035)^{6}$$ Evaluating the power: $$1.035^{6} = 1.229255$$ Substituting: $$FV = 5000 \times 1.229255$$ $$FV = 6146.28$$ $$\boxed{FV \approx \$6146.28}$$

Since the van loses 12% of its value each year, its value each year is 88% of the previous year's value, giving the depreciation (compound decrease) model $FV = PV\left(1-\dfrac{r}{100}\right)^{n}$. This can also be solved directly with the GDC finance/TVM solver using $I\% = -12$. Substituting $PV = 32\,000$, $r = 12$, $n = 5$: $$FV = 32\,000(1-0.12)^{5} = 32\,000(0.88)^{5}$$ Evaluating the power: $$0.88^{5} = 0.527732$$ $$FV = 32\,000 \times 0.527732$$ $$FV = 16\,887.42$$ $$\boxed{FV \approx \$16\,887}$$

(a) Using $FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}$ where $k=4$ (quarterly compounding) and $n=4$ years (or the GDC finance solver with $N=16$, $I\%=6$, $P/Y=1$, $C/Y=4$): $$FV = 8000\left(1+\frac{0.06}{4}\right)^{4\times4} = 8000(1.015)^{16}$$ Evaluating the power: $$1.015^{16} = 1.268986$$ $$FV = 8000 \times 1.268986$$ $$FV = 10\,151.90$$ $$\boxed{FV \approx \$10\,151.90}$$ (b) The effective annual rate satisfies $1 + r_{eff} = \left(1+\dfrac{r}{100k}\right)^{k}$, comparing one year of quarterly compounding to one annual step. $$1 + r_{eff} = (1.015)^{4}$$ $$1 + r_{eff} = 1.061364$$ $$r_{eff} = 0.061364 = 6.1364\%$$ $$\boxed{r_{eff} \approx 6.14\% \text{ (3 s.f.)}}$$

Using the compound interest formula with monthly compounding, $FV = PV\left(1+\dfrac{r}{1200}\right)^{12n}$, with $n=8$ so $12n = 96$ months (this can be solved directly using the GDC finance/TVM solver with $N=96$, $PV=-12000$, $FV=17500$, $P/Y=C/Y=12$, solving for $I\%$, or algebraically as below): $$17\,500 = 12\,000\left(1+\frac{r}{1200}\right)^{96}$$ Dividing both sides by 12 000: $$\left(1+\frac{r}{1200}\right)^{96} = \frac{17\,500}{12\,000} = 1.458333$$ Taking the 96th root of both sides (or taking $\ln$ of both sides and dividing by 96): $$1+\frac{r}{1200} = 1.458333^{1/96} = 1.003938$$ $$\frac{r}{1200} = 0.003938$$ $$r = 1200 \times 0.003938 = 4.7256$$ $$\boxed{r \approx 4.73\% \text{ (3 s.f.)}}$$

Using compound growth/decay to write expressions for both values after $n$ years, then forming and solving an inequality with logarithms (or a GDC table): $$\text{Equipment value: } E_n = 45\,000(0.85)^{n}$$ $$\text{Savings value: } S_n = 20\,000(1.09)^{n}$$ Requiring $S_n > E_n$: $$20\,000(1.09)^{n} > 45\,000(0.85)^{n}$$ Dividing both sides by $20\,000(0.85)^n$: $$\left(\frac{1.09}{0.85}\right)^{n} > \frac{45\,000}{20\,000} = 2.25$$ $$(1.282353)^{n} > 2.25$$ Taking $\ln$ of both sides: $$n > \frac{\ln 2.25}{\ln 1.282353} = \frac{0.810930}{0.248691} = 3.2608\ldots$$ Since $n$ must be a whole number, check $n=3$ and $n=4$ using a GDC: $$n=3: \quad E_3 = 45\,000(0.85)^{3} = 27\,635.63; \quad S_3 = 20\,000(1.09)^{3} = 25\,900.58$$ Savings has not yet exceeded equipment at $n=3$. $$n=4: \quad E_4 = 45\,000(0.85)^{4} = 23\,490.28; \quad S_4 = 20\,000(1.09)^{4} = 28\,231.63$$ Savings now exceeds equipment at $n=4$. $$\boxed{n = 4 \text{ complete years}}$$

(a) Using the GDC's finance (TVM) solver with: $$N = 4 \times 12 = 48 \text{ (number of payments)}, \quad I\% = 6, \quad PV = 8000, \quad FV = 0, \quad P/Y = C/Y = 12$$ Solving for $PMT$ gives: $$PMT \approx -187.88$$ $$\boxed{\text{Monthly repayment} \approx \$187.88}$$ (b) Using total repaid $=$ monthly payment $\times$ number of payments: $$\text{Total repaid} = 187.88 \times 48$$ $$\text{Total repaid} = 9018.24$$ $$\boxed{\text{Total repaid} \approx \$9018 \text{ (or } \$9020 \text{ to 3 s.f.)}}$$

Using the GDC's finance (TVM) solver with: $$N = 5 \times 12 = 60 \text{ (number of deposits)}, \quad I\% = 3.6, \quad PV = 0, \quad PMT = -250, \quad P/Y = C/Y = 12$$ Solving for $FV$ gives: $$FV \approx 16\,407.6$$ Alternatively, using the future value of an ordinary annuity formula, $FV = PMT \times \dfrac{(1+i)^N - 1}{i}$, with $i = \dfrac{0.036}{12} = 0.003$: $$FV = 250 \times \frac{(1.003)^{60} - 1}{0.003}$$ Substituting $(1.003)^{60} = 1.19690$: $$FV = 250 \times \frac{0.19690}{0.003} = 250 \times 65.632$$ $$FV = 16\,408.0$$ $$\boxed{FV \approx \$16\,408}$$

(a) Using the GDC's finance (TVM) solver with: $$N = 3 \times 12 = 36, \quad I\% = 7.2, \quad PV = 15\,000, \quad FV = 0, \quad P/Y = C/Y = 12$$ Solving for $PMT$ gives: $$PMT \approx -464.53$$ $$\boxed{\text{Monthly repayment} \approx \$464.53}$$ (b) Using total interest $=$ total repaid $-$ amount borrowed: $$\text{Total amount repaid} = 464.53 \times 36 = 16\,723.08$$ $$\text{Total interest paid} = 16\,723.08 - 15\,000$$ $$\text{Total interest paid} = 1723.08$$ $$\boxed{\text{Total interest paid} \approx \$1723 \text{ (or } \$1720 \text{ to 3 s.f.)}}$$

(a) Using the GDC's finance (TVM) solver with: $$N = 5 \times 12 = 60, \quad I\% = 5.4, \quad PV = 20\,000, \quad FV = 0, \quad P/Y = C/Y = 12$$ Solving for $PMT$ gives: $$PMT \approx -381.10$$ $$\boxed{\text{Monthly payment} \approx \$381.10}$$ (b) The outstanding balance after 24 payments equals the present value of the remaining 36 payments, evaluated at the same monthly rate. Using the GDC's amortization (balance) feature, or equivalently the TVM solver with $N = 36$ (remaining payments), $I\% = 5.4$, $PMT = -381.10$, $FV = 0$, solving for $PV$: $$\text{Balance} = PMT \times \frac{1 - (1+i)^{-36}}{i}, \qquad i = \frac{0.054}{12} = 0.0045$$ Substituting $(1.0045)^{36} = 1.17543$, so $(1.0045)^{-36} = 0.85075$: $$\text{Balance} = 381.10 \times \frac{1 - 0.85075}{0.0045} = 381.10 \times \frac{0.14925}{0.0045}$$ $$\text{Balance} = 381.10 \times 33.167 \approx 12\,640$$ $$\boxed{\text{Outstanding balance after 24 payments} \approx \$12\,640 \text{ (or } \$12\,600 \text{ to 3 s.f.)}}$$

(a) Using the future value of an ordinary annuity, $FV = PMT \times \dfrac{(1+i)^N - 1}{i}$, with $i = \dfrac{0.048}{12} = 0.004$, $PMT = 600$. Using the GDC's finance (TVM) solver with $PV = 0$, $PMT = -600$, $FV = 50\,000$, $I\% = 4.8$, $P/Y = C/Y = 12$, solving for $N$ gives $N \approx 72.1$, which is not a whole number of payments, so check whole-number values either side. $$N = 72: \quad FV = 600 \times \frac{(1.004)^{72} - 1}{0.004}$$ Substituting $(1.004)^{72} = 1.33300$: $$FV = 600 \times \frac{0.33300}{0.004} = 600 \times 83.250 = 49\,950 \text{ (just under \$50\,000)}$$ $$N = 73: \quad FV = 600 \times \frac{(1.004)^{73} - 1}{0.004}$$ Substituting $(1.004)^{73} = 1.33833$: $$FV = 600 \times \frac{0.33833}{0.004} = 600 \times 84.583 = 50\,750 \text{ (exceeds \$50\,000)}$$ Since the balance first reaches/exceeds \$50 000 after the 73rd deposit: $$\boxed{\text{It takes 73 months}}$$ (b) Using total deposited $=$ monthly deposit $\times$ number of months, and total interest $=$ balance $-$ total deposited: $$\text{Total deposited} = 600 \times 73 = 43\,800$$ Using the balance after 73 months found in part (a), $\approx \$50\,750$: $$\text{Total interest earned} = 50\,750 - 43\,800$$ $$\boxed{\text{Total interest earned} \approx \$6950}$$

Using technology (the simultaneous equation, graphing, or intersection solver) on the system: $$C = 25 + 2d$$ $$C = 15 + 3.5d$$ Setting the two expressions for $C$ equal: $$25 + 2d = 15 + 3.5d$$ Collecting terms: $$25 - 15 = 3.5d - 2d$$ $$10 = 1.5d$$ Solving for $d$: $$d = \frac{10}{1.5} = 6.6667\ldots$$ Substituting back into Plan A to find $C$: $$C = 25 + 2(6.6667) = 25 + 13.333$$ $$C = 38.333\ldots$$ $$\boxed{d \approx 6.67 \text{ GB, at a common cost of } C \approx \$38.30 \text{ (3 s.f.)}}$$

Entering the system into the GDC's simultaneous equation solver (3 equations, 3 unknowns), with coefficient matrix and constant vector: $$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2 \\ 2 & 3 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 9 \end{pmatrix}$$ Solving using the GDC gives: $$x = -0.2, \quad y = 2.8, \quad z = 1$$ Verifying by substitution into each original equation: $$\text{Eq 1: } 3(-0.2) + 2(2.8) - 1 = -0.6 + 5.6 - 1 = 4 \checkmark$$ $$\text{Eq 2: } (-0.2) - 2.8 + 2(1) = -0.2 - 2.8 + 2 = -1 \checkmark$$ $$\text{Eq 3: } 2(-0.2) + 3(2.8) + 1 = -0.4 + 8.4 + 1 = 9 \checkmark$$ $$\boxed{x = -0.2, \quad y = 2.8, \quad z = 1}$$

(a) Translating the given conditions into equations, with $c$, $a$, $s$ the numbers of child, adult, and senior tickets: $$\text{Total tickets: } c + a + s = 130$$ $$\text{Adult/senior relationship: } a = 2s + 10, \quad \text{i.e.} \quad a - 2s = 10$$ $$\text{Revenue: } 6c + 12a + 8s = 980$$ $$\boxed{\begin{cases} c+a+s=130 \\ a-2s=10 \\ 6c+12a+8s=980 \end{cases}}$$ (b) Entering this system into the GDC's simultaneous equation solver (writing the second equation as $0c + a - 2s = 10$): $$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -2 \\ 6 & 12 & 8 \end{pmatrix}\begin{pmatrix} c \\ a \\ s \end{pmatrix} = \begin{pmatrix} 130 \\ 10 \\ 980 \end{pmatrix}$$ Solving using the GDC gives: $$c = 90, \quad a = 30, \quad s = 10$$ Verifying: $90+30+10=130 \checkmark$; $2(10)+10=30=a \checkmark$; $6(90)+12(30)+8(10) = 540+360+80=980 \checkmark$ $$\boxed{\text{90 child tickets, 30 adult tickets, and 10 senior tickets were sold.}}$$

(a) Using technology, set $P(x) = Q(x)$: $$-2x^2 + 20x - 32 = 3x - 8$$ Collecting all terms on one side: $$-2x^2 + 20x - 32 - 3x + 8 = 0$$ $$-2x^2 + 17x - 24 = 0$$ Using the GDC's polynomial equation solver (or graphing the two functions and finding their points of intersection): $$x = 1.788 \text{ or } x = 6.712 \text{ (3 s.f.)}$$ Both values lie within the domain $[0,10]$. $$\boxed{x \approx 1.79 \text{ or } x \approx 6.71 \text{ (hundred units)}}$$ (b) Substituting each value into $Q(x) = 3x - 8$ (in thousands of dollars): $$x = 1.788: \quad Q = 3(1.788) - 8 = 5.363 - 8 = -2.637$$ i.e. a loss of approximately \$2640. $$x = 6.712: \quad Q = 3(6.712) - 8 = 20.137 - 8 = 12.137$$ i.e. a profit of approximately \$12 100. (Check using $P(x)$: $P(1.788) \approx -2.64$; $P(6.712) \approx 12.14$, consistent with $Q(x)$.) $$\boxed{\text{At } x \approx 1.79 \text{, a loss of about \$2640; at } x \approx 6.71 \text{, a profit of about \$12\,100.}}$$

(a) Setting $h(t) = 0$: $$-t^3 + 6t^2 + 4t = 0$$ Factoring out $t$: $$t(-t^2 + 6t + 4) = 0$$ So $t=0$ or $-t^2+6t+4=0$, i.e. $t^2 - 6t - 4 = 0$. Using the GDC's polynomial equation solver (or graphing $h(t)$ and finding its zero) on $t^2-6t-4=0$: $$t = 6.606 \text{ or } t = -0.606 \text{ (3 s.f.)}$$ Rejecting the negative solution (outside the domain), and noting $t=0$ is the launch point: $$\boxed{t \approx 6.61 \text{ seconds}}$$ (b) To find the maximum height, first differentiate: $$h'(t) = -3t^2 + 12t + 4$$ Setting $h'(t) = 0$ and using the GDC's polynomial equation solver on $3t^2 - 12t - 4 = 0$: $$t = 4.309 \text{ or } t = -0.309 \text{ (3 s.f.)}$$ Rejecting the negative solution (outside the domain): $$t \approx 4.31 \text{ seconds}$$ Substituting back into $h(t)$: $$h(4.309) = -(4.309)^3 + 6(4.309)^2 + 4(4.309)$$ $$= -80.03 + 111.42 + 17.24$$ $$= 48.63$$ $$\boxed{\text{Maximum height} \approx 48.6 \text{ m, reached at } t \approx 4.31 \text{ seconds}}$$