1. Problem (a): Find the range within which the exact value of $\frac{x}{y \cdot z}$ lies when $x=4.8$, $y=4.905$, $z=2$ rounded to nearest decimal places. Step 1. Identify the rounding intervals: - $x$ rounded to nearest 0.1: exact $x \in [4.75, 4.85)$ - $y$ rounded to nearest 0.001: exact $y \in [4.9045, 4.9055)$ - $z$ rounded to nearest integer: exact $z \in [1.5, 2.5)$ Step 2. Compute the minimum and maximum values of $\frac{x}{y\cdot z}$ using interval endpoints: - Minimum value: $$\min = \frac{\min x}{\max y \times \max z} = \frac{4.75}{4.9055 \times 2.5} \approx \frac{4.75}{12.26375} \approx 0.3873$$ - Maximum value: $$\max = \frac{\max x}{\min y \times \min z} = \frac{4.85}{4.9045 \times 1.5} \approx \frac{4.85}{7.35675} \approx 0.6596$$ So the exact value lies in $[0.3873, 0.6596]$. 2. Problem (b): Given $p$ and $q$ rounded with errors $e_p$ and $e_q$, show max relative error in $\sqrt{pq}$ is $\frac{1}{2}(|\frac{e_p}{p}| + |\frac{e_q}{q}|)$, then find exact interval of $\sqrt{1.20 \times 2.8}$. Step 1. Express $p$ and $q$ with errors: $$p = p_0 + e_p, \quad q = q_0 + e_q$$ Step 2. Absolute errors in product: $$pq = (p_0 + e_p)(q_0 + e_q) = p_0 q_0 + p_0 e_q + q_0 e_p + e_p e_q$$ Step 3. Approximate max relative error in $pq$ ignoring $e_p e_q$ small term: $$\frac{\Delta (pq)}{pq} \approx \left|\frac{e_p}{p_0}\right| + \left|\frac{e_q}{q_0}\right|$$ Step 4. For $\sqrt{pq}$, relative error approximately half that of $pq$: $$\frac{\Delta \sqrt{pq}}{\sqrt{pq}} = \frac{1}{2}\left(\left|\frac{e_p}{p_0}\right| + \left|\frac{e_q}{q_0}\right| \right)$$ Step 5. Use rounded values: $p=1.20$ rounded to nearest 0.01 so $e_p=\pm0.005$; $q=2.8$ rounded to nearest 0.1 so $e_q=\pm0.05$. Calculate relative errors: $$\left|\frac{e_p}{p}\right|=\frac{0.005}{1.20} \approx 0.00417$$ $$\left|\frac{e_q}{q}\right|=\frac{0.05}{2.8} \approx 0.01786$$ Step 6. Max relative error in $\sqrt{pq}$: $$\frac{1}{2} (0.00417 + 0.01786) = 0.01102$$ Step 7. Calculate nominal value: $$\sqrt{1.20 \times 2.8} = \sqrt{3.36} \approx 1.833$$ Step 8. Interval for exact value: $$[1.833(1 - 0.01102),\ 1.833(1 + 0.01102)] = [1.812, 1.853]$$ 3. Physics problem (a): Particle of mass 4 kg, force $\vec{F} = (4\hat{i} + 12t\hat{j} - 3\hat{k})$ N acting, find acceleration $\vec{a}(t)$. Step 1. Newton's 2nd law: $\vec{F} = m \vec{a}$. Step 2. Acceleration: $$\vec{a}(t) = \frac{\vec{F}}{m} = \frac{1}{4} (4\hat{i} + 12t\hat{j} - 3\hat{k}) = (1\hat{i} + 3t\hat{j} - 0.75\hat{k})\ \text{m/s}^2$$ 4. Physics problem (b): Velocity $\vec{v}(t)$ given particle starts from rest $\vec{v}(0)=0$. Step 1. Velocity is integral of acceleration: $$\vec{v}(t) = \int \vec{a}(t) dt = \int (1\hat{i} + 3t\hat{j} - 0.75\hat{k}) dt = (t\hat{i} + \frac{3t^2}{2}\hat{j} - 0.75t\hat{k}) + \vec{C}$$ Given $\vec{v}(0) = 0$ implies $\vec{C} = 0$, so: $$\vec{v}(t) = (t\hat{i} + \frac{3t^2}{2}\hat{j} - 0.75t\hat{k})$$ 5. Physics problem (c): Work done by force $\vec{F}$ after 2 seconds. Step 1. Displacement $\vec{s}$ from initial velocity zero: $$\vec{s}(t) = \int \vec{v}(t) dt = \int (t\hat{i} + \frac{3t^2}{2}\hat{j} - 0.75t\hat{k}) dt = (\frac{t^2}{2}\hat{i} + \frac{t^3}{2}\hat{j} - \frac{0.75 t^2}{2}\hat{k}) + \vec{D}$$ With $\vec{s}(0)=0$, $\vec{D}=0$: $$\vec{s}(t) = \left(\frac{t^2}{2}\hat{i} + \frac{t^3}{2}\hat{j} - \frac{0.375 t^2}{1}\hat{k}\right)$$ Step 2. Calculate displacement at $t=2$ s: $$\vec{s}(2) = \left(\frac{4}{2}\hat{i} + \frac{8}{2}\hat{j} - 0.375 \times 4 \hat{k} \right) = (2\hat{i} + 4\hat{j} -1.5\hat{k})$$ Step 3. Work done $W = \vec{F} \cdot \vec{s}$, with average force over time interval or integrate force times velocity: Work done: $$W = \int_0^2 \vec{F} \cdot \vec{v} dt$$ Calculate: $$\vec{F} \cdot \vec{v} = (4)(t) + (12t)(\frac{3t^2}{2}) + (-3)(-0.75 t) = 4t + 18 t^3 + 2.25 t = 6.25 t + 18 t^3$$ Integrate over 0 to 2: $$W = \int_0^2 (6.25 t + 18 t^3) dt = \left[3.125 t^2 + 4.5 t^4 \right]_0^2 = 3.125 \times 4 + 4.5 \times 16 = 12.5 + 72 = 84.5\ \text{J}$$ 6. Statistics problem (i): Normal variable $X$, $P(X < 35) = 0.2$, $P(35 < X < 45) = 0.65$, find mean $\mu$ and std dev $\sigma$. Step 1. Use standard normal $Z = \frac{X - \mu}{\sigma}$ Step 2. From $P(X<35)=0.2$, $Z_1 = \frac{35 - \mu}{\sigma} = z_{0.2} \approx -0.8416$ Step 3. From $P(35 40)$. Step 1. Convert to $Z$: $$Z = \frac{40 - 39.49}{5.33} = \frac{0.51}{5.33} = 0.096$$ Step 2. $P(X > 40) = P(Z > 0.096) = 1 - P(Z < 0.096) \approx 1 - 0.5383 = 0.4617$ 8. Statistics problem: Mean age calculation from class intervals and frequencies. Step 1. Class midpoints: - [18,19): 18.5 - [19,20): 19.5 - [20,24): 22 - [24,26): 25 - [26,30): 28 - [30,32): 31 Step 2. Frequencies: 24, 70, 76, 48, 16, 6 Step 3. Calculate weighted sum: $$ \sum f x = 24\times18.5 + 70\times19.5 + 76\times22 + 48\times25 + 16\times28 + 6\times31 $$ $$ = 444 + 1365 + 1672 + 1200 + 448 + 186 = 5315$$ Step 4. Total frequency: $$ N = 24 + 70 + 76 + 48 + 16 + 6 = 240$$ Step 5. Mean: $$\bar{x} = \frac{5315}{240} \approx 22.15\ \text{years}$$ 9. Histogram and modal age estimation omitted as no graph requested. Final answers summary: - (a) $\frac{x}{y z} \in [0.3873, 0.6596]$ - (b) Max relative error in $\sqrt{pq} = \frac{1}{2}(\left|\frac{e_p}{p}\right|+\left|\frac{e_q}{q}\right|)$; interval of $\sqrt{1.20\times 2.8} = [1.812, 1.853]$ - Physics (a) $\vec{a}(t) = (1, 3t, -0.75)$ - Physics (b) $\vec{v}(t) = (t, \frac{3t^2}{2}, -0.75 t)$ - Physics (c) Work done after 2s = 84.5 J - Stats (i) $\mu \approx 39.49$, $\sigma \approx 5.33$ - Stats (ii) $P(X>40) \approx 0.462$ - Mean student age $\approx 22.15$ years