Suppose both the mean and median of the distribution are 6. which of these statements is true about the mode of the distribution

for one thing their called fish not "fishes" and 2 fish can't be dead and alive and this question is absurd.

step-by-step explanation:

the value of x is 90 - 20 aka 70*, these are complementary angles

step-by-step explanation:

by algebra, we combine the fractions: c + 1/c = c²/c + 1/c = (c²+1)/c ≥ 2.

(c²+1)/c ≥ 2 and we multiply both sides by c. c is positive so inequality sign stay s.

c² + 1 ≥ 2c ⇔ c²-2c + 1 = (c - 1)(c - 1) ≥ 0

consider case where 0 < c < 1. by the subtraction property of equality, subtracting through by 1, we have

-1 < c - 1 < 0

so c - 1 is negative for 0 < c < 1. but (c - 1)(c - 1)   means that we have a negative times a negative which is a positive. thus (c - 1)(c - 1) ≥ 0 is true for 0 < c < 1.

consider case where c = 1. then (1-1)(1-1) = 0 ≥ 0 which is true

consider the case where c > 1. by the subtraction property of equality, this is equivalent to c - 1 > 0. so c - 1 is a positive number, and (c-1)(c-1) is positive, thereby satisfying (c - 1)(c - 1) ≥ 0.

so c + 1/c ≥ 2 is true for all c > 0.