Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
12
They all equal each other. a = b = c = d = e e = a e = b e = c e = d e = e
a d e e c b a e d d a c
12=n in a d
12 eggs in a dozen
12 eggs in a dozen
12 eggs is a dozen
12 dozens in a carton?
E. D. Smith was born on 1853-12-08.
12=eggs in a dozen
12 eggs in a dozen.
12 eggs in a dozen
Assuming by "numbers" you mean "whole numbers":{10, 10, 12, 13, 15}If the five numbers are {a, b, c, d, e} with a ≤ b ≤ c ≤ d ≤ e, then:median = 12 → c = 12leaving only two numbers (a, b) ≤ 12.mode = 10 → two or more numbers equal 10 (which is less than 12) → a = b = 10The final two numbers (d, e) are not equal, both greater than 12, and such that the sum of all five numbers is 60.10 + 10 + 12 + d + e = 60 → d + e = 28 → d = 13, e = 15→ {a, b, c, d, e} = {10, 10, 12, 13, 15}[If "numbers" includes "decimal numbers" (ie numbers with a fractional part), then as long as d + e = 28 and 12 < d < e there are infinitely many solutions, eg {10, 10, 12, 12.5, 15.5}, {10, 10, 12, 13.75, 14.25}]
eggs in a dozen!
For a hint, they are usually scrambled, boiled, or fried and served with breakfast. The E is the first letter of what they are, and the 12 and the D are different ways of saying how many there are.
#f #e #e#e#d#d#d#d#e#d#e#d#e#d#e#d#e#d#e#d