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If the percentage of adenine (A) in human DNA is 32%, the percentage of thymine (T) would also be 32%. This calculation is based on Chargaff's rules, which state that in a double-stranded DNA molecule, nitrogenous bases always pair in a specific manner: Adenine (A) always pairs with Thymine (T) (A = T). Guanine (G) always pairs with Cytosine (C) (G = C)

Both statements in options B and C are not correct based on the biological principles of sex-linked inheritance: Males cannot be carriers: Because colour blindness is an X-linked recessive trait and males have only one X chromosome, they will always express the condition if they possess the gene. They cannot "carry" it without being affected. Females can be carriers: Females have two X chromosomes. If they have the gene on only one of these chromosomes, they do not express the condition but can pass it to their offspring.

Lamarck's theory, also known as Lamarckism or the inheritance of acquired characteristics, posits that physical changes an organism undergoes during its lifetime—such as the development of an organ through increased use or its weakening through disuse—can be passed directly to its offspring. He famously used the example of giraffes stretching their necks to reach high leaves, suggesting this stretching resulted in longer necks that were then inherited by the next generation.

The theory of natural selection was formulated by Charles Darwin, who independently reached the same conclusion.

\(\frac{16^{2 + x}}{4} = 64^x\) 16\(^{2 + x} = 4 \times 64^x\) 4\(^{2(2 + x)} = 4^1 \times 4^{3(x)}\) (Equate coefficients since the bases are the same) 2(2 + x) = 1 + 3x 4 + 2x = 1 + 3x 2x - 3x = 1 - 4 -x = -3 x = 3.

Length of Arc = \(\frac{\theta}{360}\) x 2\(\pi\)r Length of arc = 33cm, diameter = 28cm → r = 14cm 33 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7}\times14\) \(\theta\) = \(\frac{33 \times 180}{22 \times 2}\) \(\theta\) = 135º

Length of a chord = 2rSin\(\frac{\theta}{2}\) \(\theta\) = 60º not 30º and radius = 9cm Therefore, Length of the chord = 2 x 9 Sin\(\frac{60º}{2}\) = 2 x 9 x Sin 30º = 2 x 9 x \(\frac{1}{2}\) = 9cm.

The area of a triangle given two sides and the included angle = \(\frac{1}{2}\) x a x b x Sin\(\theta\) a = b = 8cm \(\theta\) = 60º Area = \(\frac{1}{2}\) x 8 x 8 x Sin 60º = \(\frac{1}{2}\) x 64 x \(\frac{\sqrt{3}}{2}\) ( Sin 60º = \(\frac{\sqrt{3}}{2}\) ) = \(\frac{64\sqrt{3}}{4}\) = 16\(\sqrt{3}cm^2\)

The construction of angle 30º (If you take a 60º angle and bisect it, you would divide it into two angles of 30º each.)

\(\overline{MN} || \overline{KL}\), \(\triangle\) LXK is similar to \(\triangle\)MXN So, the ratio of the areas of the two similar triangles equals the square of the ratio of their corresponding sides Therefore, \(\frac{ Area of \triangle MXN}{Area of \triangle LXK} = \frac{12^2}{9^2}\) \(\frac{16}{Area of \triangle LXK} = \frac{144}{81}\) Area of \(\triangle\)LXK = \(\frac{ 16 \times 81}{144} = 9cm^2\)

10111.11\(_{two}\) convert the number to base 10. 10111.11\(_{two}\) = 1 x 2\(^4\) + 0 x 2\(^3\) + 1 x \(^2\) + 1 x 2\(^1\) + 1 x 2\(^0\) + 1 x 2\(^{-1}\) + 1 x 2\(^{-2}\) = 16 + 0 + 4 + 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) = 23 + \(\frac{3}{4}\) = 23\(\frac{3}{4}\)

Subtract 14256\(_{seven}\) from 20045\(_{seven}\) = 2456\(_{seven}\) 2 0 0 4 5 1 4 2 5 6 0 2 4 5 6

16\(^x\) = 0.25 16\(^x\) = \(\frac{25}{100} = \frac{1}{4}\) 4(\(^2)^x\) = 4\(^{-1}\) 2x = -1 x = \(\frac{-1}{2}\)

\(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) multiply the denominator and the numerator by the conjugate of 3 + \(\sqrt{7}\) → 3 - \(\sqrt{7}\) \(\frac{ 5 + \sqrt{7}}{3 + \sqrt{7}}\) x \(\frac{3 - \sqrt{7}}{3 - \sqrt{7}}\) \(\frac{15 - 5\sqrt{7} + 3\sqrt{7} - 7}{3^2 - (\sqrt{7})^2}\) \(\frac{8 - 2\sqrt{7}}{9 - 7}\) = \(\frac{ 8 - 2\sqrt{7}}{2}\) = 4 - \(\sqrt{7}\)

65 = 30 -x + x + 45 - x 65 - 75 = 10 = x ( number of students offering both subjects) Pr( \(\text{students offering both subjects}\)) = \(\frac{10}{65}\) = \(\frac{2}{13}\)