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### Question 3 # Explanation: ## Step1: Recall decibel formula The decibel level $L$ of a sound with intensity $I$ is given by $L = 10\log\left(\frac{I}{I_0}\right)$. Let $I_1$ and $I_2$ be the intensities of the two sounds with $L_1$ and $L_2$ as their decibel - levels respectively, and $\frac{I_1}{I_2}=647$. We know $L_1 = 10\log\left(\frac{I_1}{I_0}\right)=97$ and $L_2 = 10\log\left(\frac{I_2}{I_0}\right)$. ## Step2: Express the difference in decibel levels $L_1 - L_2=10\log\left(\frac{I_1}{I_0}\right)-10\log\left(\frac{I_2}{I_0}\right)$. Using the logarithm property $\log a-\log b=\log\frac{a}{b}$, we get $L_1 - L_2 = 10\log\left(\frac{\frac{I_1}{I_0}}{\frac{I_2}{I_0}}\right)=10\log\left(\frac{I_1}{I_2}\right)$. ## Step3: Calculate the difference Since $\frac{I_1}{I_2}=647$, then $10\log\left(\frac{I_1}{I_2}\right)=10\log(647)$. $\log(647)\approx 2.81$, so $10\log(647)\approx 28.1$. ## Step4: Find $L_2$ $L_2 = L_1-10\log\left(\frac{I_1}{I_2}\right)$. Substituting $L_1 = 97$ and $10\log\left(\frac{I_1}{I_2}\right)\approx 28.1$, we get $L_2=97 - 28.1=68.9$ dB. # Answer: 68.9 dB ### Question 4 # Explanation: ## Step1: Simplify each part of the fraction First, $2^6=(2^2)^3 = 4^3$. Second, $\left(\frac{1}{4}\right)^5 = 4^{- 5}$. Third, $\sqrt[4]{16}=2$, so $(\sqrt[4]{16})^3=2^3=(2^2)^{\frac{3}{2}} = 4^{\frac{3}{2}}$. ## Step2: Rewrite the fraction with base - 4 The original expression $\frac{2^6\times\left(\frac{1}{4}\right)^5}{(\sqrt[4]{16})^3}$ becomes $\frac{4^3\times4^{-5}}{4^{\frac{3}{2}}}$. ## Step3: Use the exponent rule $a^m\times a^n=a^{m + n}$ and $\frac{a^m}{a^n}=a^{m - n}$ $4^3\times4^{-5}=4^{3+( - 5)}=4^{-2}$. Then $\frac{4^{-2}}{4^{\frac{3}{2}}}=4^{-2-\frac{3}{2}}=4^{-\frac{4 + 3}{2}}=4^{-\frac{7}{2}}$. # Answer: $4^{-\frac{7}{2}}$