सविस्तर स्पष्टीकरण (Detailed Explanation):
अचूक उत्तर: पर्याय (3)
दिलेले आहे:
निरीक्षणांचे मूल्य: 16, 15, 7, 5, 4, 2, 7 आणि 8.
वापरलेले सूत्र:
सरासरी = सर्व 'n' निरीक्षणांची बेरीज ÷ 'n'
(a + b)2 = a2 + b2 + 2ab
(a2 - b2) = (a + b)(a - b)
गणना:
16, 15, 7, 5, 4, 2, 7 आणि 8 या निरीक्षणांचे सरासरी मूल्य = a
आता,
⇒ a = \(\frac{16+15+7+5+4+2+7+8}{8}\)
⇒ a = \(\frac{64}{8}\)
⇒ a = 8
आता, संबंधित चलाचे मूल्य आवश्यक पदावलीमध्ये टाका
⇒ \(\frac{3+√{8}}{3-√{8}}\)
आता, छेदाचे परीमेयीकरण करा.
⇒ \(\frac{3+√{8}}{3-√{8}} \times \frac{3+√{8}}{3+√{8}}\)
⇒ \(\frac{(3+√{8})^2}{3^2-√{8}^2} \)
⇒ \(\frac{(9 + 8 + 6 √{8})}{9-8} \)
⇒ \(17 + 6\sqrt{4 \times 2}\)
⇒ 17 + 12√2
∴ पदावलीचे मूल्य 17 + 12√2 आहे.
Given:
The value of observations: 16, 15, 7, 5, 4, 2, 7 and 8.
Formula used:
Average = Sum of all 'n' observations ÷ 'n'
(a + b)^(2) = a^(2) + b^(2) + 2ab
(a^(2) - b^(2)) = (a + b)(a - b)
Calculation:
The average value of the given observations 16, 15, 7, 5, 4, 2, 7 and 8 = a
Now,
⇒ a = \(\frac{16+15+7+5+4+2+7+8}{8}\)
⇒ a = \(\frac{64}{8}\)
⇒ a = 8
Now, substitute the value of the respective variable in the required expression.
⇒ \(\frac{3+√{8}}{3-√{8}}\)
Now, perform the rationalisation to the denominator.
⇒ \(\frac{3+√{8}}{3-√{8}} \times \frac{3+√{8}}{3+√{8}}\)
⇒ \(\frac{(3+√{8})^2}{3^2-√{8}^2} \)
⇒ \(\frac{(9 + 8 + 6 √{8})}{9-8} \)
⇒ \(17 + 6\sqrt{4 \times 2}\)
⇒ 17 + 12√2
∴ The value of the required expression is 17 + 12√2 .