The trailing zero after the decimal point counts as a significant figure.

All non-zero digits are significant, so 1002 has 4 significant figures.

5000 has 1 significant figure unless specified otherwise (e.g., 5000. has 4 significant figures).

5000 has 1 significant figure unless specified with a decimal point (e.g., 5000. has 4).

The number of subsets of a set with n elements is 2^n. Here, n = 4, so 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 6, so the number of subsets is 2^6 = 64.

The number of subsets of a set with n elements is 2^n. Here, n = 6, so 2^6 = 64.

The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 6, so the number of subsets is 2^6 = 64.

The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.

The number of subsets of a set with n elements is given by 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

The number of subsets of a set with n elements is 2^n. Here, n=4, so the number of subsets is 2^4 = 16.

The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.

The only subset that does not contain y is {∅} and {x}. Total = 2.

The Earth is 360 degrees, and if each time zone is 15 degrees, then 360 / 15 = 24 time zones.

The number of ways = 5C2 * 4C2 = 10 * 6 = 60.

The number of ways to select 2 boys from 5 is C(5, 2) and 2 girls from 5 is C(5, 2). Total = C(5, 2) * C(5, 2) = 10 * 10 = 100.

The number of ways is C(6,2) * C(4,2) = 15 * 6 = 90.

The number of ways is 5C2 * 6C3 = 10 * 20 = 200.

The number of ways is 5C2 * 7C3 = 10 * 35 = 350.

The number of ways is 6C2 * 4C3 = 15 * 4 = 60.

The number of ways is 8C2 * 6C3 = 28 * 20 = 560.

The number of ways to select 2 boys from 5 is 5C2 and 3 girls from 7 is 7C3. Total = 5C2 * 7C3 = 10 * 35 = 350.

The number of ways to select 2 students from 5 is given by 5C2 = 10.

The number of ways to select 2 from 8 is C(8, 2) = 28.

The number of ways to select 3 books from 10 is given by 10C3 = 120.

The number of ways to arrange 5 people is 5! = 120.

The number of ways to choose 3 books from 7 is 7C3 = 7! / (3! * 4!) = 35.

The number of ways to choose 3 from 7 is C(7, 3) = 35.