The question involves calculating the percentage of policyholders that will renew at least one policy next year based on the given probabilities of renewal and the distribution of policy types among policyholders. The insurance company provides the following probabilities:

1. 40% of policyholders with only an auto policy will renew next year.
2. 70% of policyholders with only a homeowner policy will renew next year.
3. 80% of policyholders with both an auto and a homeowner policy will renew at least one policy next year.

Additionally, the company records show the following distribution of policy types:

1. 70% of policyholders have an auto policy.
2. 50% of policyholders have a homeowner policy.
3. 20% of policyholders have both an auto and a homeowner policy.

To find the percentage of policyholders that will renew at least one policy, we use the law of total probability. We define the following events:

• A: A policyholder has an auto policy.
• H: A policyholder has a homeowner policy.

From the given information, we calculate the probabilities of the intersections and complements of A and H:

• P(A ∩ H) = 0.20 (20% have both policies)
• P(A ∩ H') = P(A) - P(A ∩ H) = 0.70 - 0.20 = 0.50 (50% have only an auto policy)
• P(A' ∩ H) = P(H) - P(A ∩ H) = 0.50 - 0.20 = 0.30 (30% have only a homeowner policy)

Now, we calculate the proportion of policyholders that will renew at least one policy by considering all possible scenarios:

• 0.40 * P(A ∩ H') = 0.40 * 0.50 = 0.20 (40% of those with only an auto policy will renew)
• 0.70 * P(A' ∩ H) = 0.70 * 0.30 = 0.21 (70% of those with only a homeowner policy will renew)
• 0.80 * P(A ∩ H) = 0.80 * 0.20 = 0.16 (80% of those with both policies will renew at least one)

Adding these proportions together gives us the total percentage of policyholders that will renew at least one policy:

0.20 + 0.21 + 0.16 = 0.57

Therefore, 57% of policyholders will renew at least one policy next year, which corresponds to option D.