answer:Since n in mathbb{N}^* and 3C_{n-1}^{n-5} = 5A_{n-2}^2, We have 3 cdot frac{(n-1)(n-2)(n-3)(n-4)}{4 times 3 times 2 times 1} = 5 cdot (n-2)(n-3), This simplifies to (n-1)(n-4)=40, Rearranging, we get n^2 - 5n - 36 = 0. Solving this quadratic equation, we get n=9 or n=-4 (which is not applicable to this problem, so we discard it). Therefore, the value of n is boxed{9}. Hence, the answer is B. This problem involves the application of combinations and permutations formulas to find the value of n. By listing out the equations and solving for n, we can find the solution. This is a basic problem that tests the understanding of combinations and permutations formulas.

answer:If Anna wants to have 8 lettuce plants worth of salads, and each plant provides 3 large salads, then without any loss, she would have: 8 plants * 3 salads per plant = 24 large salads. However, Anna estimates that half of the lettuce will be lost to insects and rabbits. Therefore, she will only have half of the 24 large salads: 24 large salads / 2 = 12 large salads. So, Anna wants to have boxed{12} large salads.

answer:To find the value of k, you can rearrange the equation to solve for k by multiplying both sides by k and then dividing both sides by 4: 56 / k = 4 Multiply both sides by k: 56 = 4k Now divide both sides by 4: 56 / 4 = k k = 14 So, k = boxed{14} .

answer:Given the relationship ( A = 2p ) and using the formula ( A = rs ) where ( r ) is the inradius and ( s ) is the semiperimeter, we know from the problem that ( A = 2p ) and ( p = 2s ), thus ( A = 4s ). Plugging into the area formula: [ A = rs = 4s ] [ r = 4 ] Therefore, the radius of the inscribed circle is ( boxed{4} ).