Choose a course and topic, attempt each question and open its complete worked solution when you are ready.
If P(A)=\(\frac{2}{11}\), find P(A′).
Define the sample space and count each favourable outcome once.
The final value lies in [0,1]; complement pairs also sum to one, providing an independent check.
Values 6,10,14,18,24,28,30 have frequencies 2,4,7,12,8,4,3. Find variance and standard deviation.
Form the required totals explicitly and divide by the full number of observations.
The variance is nonnegative and the reported standard deviation is its positive square root.
Evaluate lim(x→0) cos x/(π−x).
Simplify the expression before substitution and preserve one-sided information.
Substitution into the original expression verifies the stated domain and final value.
Complete every requested part of exercise 5.1, question 16: ≥ − 4 3 5 3 4 5 Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line
Proceed from the given expression and keep every transformation explicit.
The last line satisfies the required domain and establishes the result.
Complete every requested part of exercise 3.2, question 9: sin (– ) 3 3 15π
Proceed from the given expression and keep every transformation explicit.
The last line satisfies the required domain and establishes the result.
Find
Substituting the six allowed \(x\)-values gives \(R=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}\).
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Without peroxide, protonation forms the more stable secondary carbocation and Br⁻ gives Markovnikov product. Peroxide initiates a radical chain in which Br• adds to create the more stable secondary radical, placing Br at the terminal carbon.
For the stated O–O, C–O, C–Br and aromatic pi-bond cleavages, show electron flow, classify homolysis or heterolysis, and name the reactive intermediates.
Single-headed arrows show one-electron movement in homolysis; double-headed arrows show electron-pair movement in heterolysis.
Why does the following reaction occur ? XeO6 – (aq) + 2F (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l) – What conclusion about the compound Na4XeO6 (of which XeO6– is a part) can be drawn from the reaction.
Xe in XeO₆⁴⁻ is +8 and is reduced to +6 in XeO₃, while F⁻ is oxidised to F₂.
Determine the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14
Let x M each of I₂ and Cl₂ form; ICl becomes 0.78−2x. Solve x²/(0.78−2x)²=0.14.