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(D) To find the approximate value,we round the numbers to the nearest integers:$24.98 \approx 25$,$16.02 \approx 16$,$7.98 \approx 8$,$15.04 \approx 1

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$20$

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(D) To find the approximate value,we round the numbers to the nearest integers:$24.98 \approx 25$,$16.02 \approx 16$,$7.98 \approx 8$,$15.04 \approx 15$,$38.93 \approx 39$.Substituting these values into the equation:$25^{2} 	imes \frac{16^{2}}{8 	imes 15} 	imes 39 = 130 	imes x^{2}$$\frac{625 	imes 256}{120} 	imes 39 = 130 	imes x^{2}$Simplify the expression:$\frac{625 	imes 256 	imes 39}{120} = 130 	imes x^{2}$$x^{2} = \frac{625 	imes 256 	imes 39}{120 	imes 130}$$x^{2} = \frac{625 	imes 256 	imes 39}{15600} = \frac{6240000}{15600} = 400$$x = \sqrt{400} = 20$.

What approximate value will come in place of the question mark $(?)$ in the given equation? (You are not expected to calculate the exact value)
$24.98^{2} \times \frac{16.02^{2}}{(7.98 \times 15.04)} \times 38.93 = 130 \times ?^{2}$

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What approximate value will come in place of the question mark $(?)$ in the given equation? (You are not expected to calculate the exact value)$24.98^{2} \times \frac{16.02^{2}}{(7.98 \times 15.04)} \times 38.93 = 130 \times ?^{2}$