answer:The answer is 32. A put option will only be exercised if the price of the stock at expiry is lower than the exercise price. When exercised, the payoff from the put option is: payoff from put option = max(0, exercise price - asset price) Thus, the lower the asset price at expiry, the higher the payoff from the put option. Among the possible market prices lower than the exercise price of 35, 32 is the highest price that would still result in a positive payoff when exercising the option (35 - 32 - 3 = 0).

answer:a) To calculate the present value of the annuity, we can use the formula: PV = M * [1 - (1 + r)^-T] / r where PV is the present value, M is the annual payment, r is the annual interest rate, and T is the number of years. Plugging in the given values, we get: PV = 14,000 * [1 - (1 + 0.11)^-9] / 0.11 PV = 14,000 * [1 - 0.4632] / 0.11 PV = 14,000 * 0.5368 / 0.11 PV = 69,836.63 Therefore, the present value of the annuity is 69,836.63. b) To calculate the worth of the annuity at the end of 9 years, we can use the formula: FV = PV * (1 + r)^T where FV is the future value, PV is the present value, r is the annual interest rate, and T is the number of years. Plugging in the given values, we get: FV = 69,836.63 * (1 + 0.11)^9 FV = 69,836.63 * 2.5604 FV = 178,644.69 Therefore, the worth of the annuity at the end of 9 years is 178,644.69.

answer:Given the vector {eq}displaystyle vec alpha =-4 overrightarrow{hat{b}} -5 vec{j}-3 vec{k} {/eq}, we need to find its magnitude, denoted as {eq}|vec{alpha}| {/eq}. The magnitude of a vector {eq}displaystyle vec a = a_1 hat i + a_2 hat j + a_3 hat k {/eq} (where {eq}hat i, hat j, hat k {/eq} are unit vectors along the x, y, and z axes, respectively) is calculated as: {eq}displaystyle | vec a | = sqrt { a_1^2 + a_2^2 + a_3^2 } {/eq} Since {eq}overrightarrow{hat{b}} {/eq} is a unit vector along the x-axis similar to {eq}hat i {/eq}, we can rewrite the vector as: {eq}vec alpha = -4 hat i - 5 hat j - 3 hat k {/eq} Now, we can calculate the magnitude: {eq}displaystyle | vec alpha | = sqrt { (-4)^2 + (-5)^2 + (-3)^2 } = sqrt { 16 + 25 + 9 } {/eq} {eq}displaystyle | vec alpha | = sqrt { 50 } {/eq} {eq}displaystyle | vec alpha | = sqrt { 25 cdot 2 } {/eq} {eq}displaystyle | vec alpha | = 5sqrt{2} {/eq} So, the magnitude of the vector {eq}vec{alpha} {/eq} is {eq}boxed{5sqrt{2}} {/eq}.

answer:To solve the system of linear equations, we can use the method of substitution or elimination. Here, we will use the elimination method. First, we add the two equations together: (10x + y + z) + (-10x + 3y + 10z) = -9 + 0 Simplifying the equation, we get: 4y + 11z = -9 Next, we multiply the first equation by 3 and the second equation by 1: 3(10x + y + z) = 3(-9) -1(10x + 3y + 10z) = -1(0) Simplifying the equations, we get: 30x + 3y + 3z = -27 -10x - 3y - 10z = 0 Now, we add the two equations together: (30x + 3y + 3z) + (-10x - 3y - 10z) = -27 + 0 Simplifying the equation, we get: 20x - 7z = -27 We now have two equations: 4y + 11z = -9 20x - 7z = -27 We can solve for z in the second equation: 20x - 7z = -27 -7z = -27 - 20x z = frac{-27 - 20x}{-7} z = frac{27 + 20x}{7} We can substitute this expression for z into the first equation: 4y + 11z = -9 4y + 11(frac{27 + 20x}{7}) = -9 4y + frac{297 + 220x}{7} = -9 28y + 297 + 220x = -63 28y + 220x = -360 We can now solve for y in this equation: 28y + 220x = -360 28y = -360 - 220x y = frac{-360 - 220x}{28} y = -frac{90 + 55x}{7} Finally, we can substitute the expressions for y and z into the first equation to solve for x: 10x + y + z = -9 10x + (-frac{90 + 55x}{7}) + (frac{27 + 20x}{7}) = -9 10x - frac{90 + 55x}{7} + frac{27 + 20x}{7} = -9 frac{70x - 90 - 55x + 27 + 20x}{7} = -9 frac{35x - 63}{7} = -9 35x - 63 = -63 35x = 0 x = 0 Therefore, the solution to the system of linear equations is x = 0, y = -frac{90 + 55(0)}{7} = -frac{90}{7}, and z = frac{27 + 20(0)}{7} = frac{27}{7}. The solution to the system of linear equations is x = 3, y = -60, and z = 21.