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The sentence "Imona couldn't have attended lectures yesterday" suggests a situation where it is unlikely or impossible for Imona to have attended lectures on the specified day. The word "couldn't" implies a strong negation or impossibility. Option A, "It was impossible that Imona attended lectures yesterday," is the correct interpretation. It reflects the idea that attending lectures yesterday is not a possible or likely scenario for Imona.

\(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 2 & 10 & -3 & 9 & 45\\ 4 - 6 & 8 & 5 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 8 & 40 & 3 & 9 & 45 \\ \hline & 18 & & 90 & & & 90 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{90}{18}\) = 5 S.D = \(\frac{\sum f(x - x)^2}{\sum f}\) = \(\frac{90}{18}\) = \(\sqrt{5}\)

Total number of students (y) using the subject French ( 90 / 360 )º x Y = 30 Y = 30 x 4 Y = 120 History- 90º + French- 90º + Economics- 150º + CRK = 360º CRK = 360º - 330º = 30º Number that offered CRK: ( 30 / 360 )º x 120 = 10

Addition of vectors: P = Q + R

Capacitors in parallel = 4\(\mu\)f + 2 \(\mu\) f = 6\(\mu\) f Capacitors in series =\(\frac{1}{C_{eff}}\) = \(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}\) \(\frac{1}{C_{eff}}\) = \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{4}\) = \(\frac{6 + 4 + 2 + 3}{12}\) \(\frac{1}{C_{eff}}\) = \(\frac{15}{12}\) C\(_{eff}\) = \(\frac{12}{15}\) = 0.8\(\mu\) f

Option B is a graph of weak a acid and strong base

The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 8 protons is a carbon atom, no matter how many neutrons may be present). If an atom loses or gains electrons, it will become a positively or negatively charged particle, called an ion. The loss of one or more electrons results in more protons than electrons and an overall positively charged ion, called a cation.

No expanded working has been attached for this question yet. The verified answer is Option D.

The carbon (IV) oxide produced along with the hydrogen is removed by dissolving it in water under a pressure of 30atm or other solvents like caustic soda solution.

The carbonyl group in tha alkanal (aldehyde) family is always at the end of a carbon chain

\(3310_{5} - 1442_{5} = 1313_{5}\) You can either do a direct subtraction (which is easier), or convert both numbers in base 10, subtract and convert back to base 5 (if you're not sure).

3.1415926 \(\approxeq\) 3.14159 (to 5 d.p)

Error = (16.8cm - 15cm = 1.8cm\) \(\text{% error} = \frac{1.8}{15} \times 100% = \frac{180}{15}\) = \(12.00%\)

Present salary = N24,000. Increment after first year = \(\frac{10}{100} \times N24,000 = N2,400\) Salary at the beginning of second year = N(24,000 + 2,400) = N26,400. Increment after second year = \(\frac{10}{100} \times N26,400 = N2,640\) Salary at the beginning of third year = N(26,400 + 2,640) = N29,040. \(\therefore\) His salary at the beginning of the third year = N29,040.

\(0.0014 = 1.4 \times 10^{-3}\) \(0.011 = 1.1 \times 10^{-2}\) \(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\) = \(1.54 \times 10^{-5}\)