1. **Problem Statement:** We need to find the equivalent noise level when five machines with noise levels A=90 dB, B=84 dB, C=97 dB, D=88 dB, and E=94 dB are all running together. 2. **Formula Used:** Noise levels in decibels (dB) are logarithmic. To combine noise levels, we convert each dB value to its corresponding intensity, sum the intensities, and then convert back to dB. The formula to convert from decibels to intensity is: $$I = 10^{\frac{L}{10}}$$ where $L$ is the noise level in dB. The combined intensity is: $$I_{total} = I_A + I_B + I_C + I_D + I_E$$ The equivalent noise level $L_{eq}$ is: $$L_{eq} = 10 \log_{10}(I_{total})$$ 3. **Calculate individual intensities:** $$I_A = 10^{\frac{90}{10}} = 10^9$$ $$I_B = 10^{\frac{84}{10}} = 10^{8.4}$$ $$I_C = 10^{\frac{97}{10}} = 10^{9.7}$$ $$I_D = 10^{\frac{88}{10}} = 10^{8.8}$$ $$I_E = 10^{\frac{94}{10}} = 10^{9.4}$$ 4. **Evaluate intensities numerically:** $$I_A = 1.0 \times 10^9$$ $$I_B \approx 2.5119 \times 10^8$$ $$I_C \approx 5.0119 \times 10^9$$ $$I_D \approx 6.3096 \times 10^8$$ $$I_E \approx 2.5119 \times 10^9$$ 5. **Sum the intensities:** $$I_{total} = 1.0 \times 10^9 + 2.5119 \times 10^8 + 5.0119 \times 10^9 + 6.3096 \times 10^8 + 2.5119 \times 10^9$$ $$I_{total} = (1.0 + 0.25119 + 5.0119 + 0.63096 + 2.5119) \times 10^9$$ $$I_{total} = 9.40695 \times 10^9$$ 6. **Calculate equivalent noise level:** $$L_{eq} = 10 \log_{10}(9.40695 \times 10^9)$$ $$= 10 (\log_{10}(9.40695) + \log_{10}(10^9))$$ $$= 10 (0.973 + 9) = 10 \times 9.973 = 99.73$$ 7. **Final answer:** The equivalent noise level when all machines are running is approximately **99.7 decibels**.