A list of 5 positive integers has all of the following properties:.
• the only integer in the list that occurs more than once is 8,
• its median is 9, and
• its average (mean) is 10.
What is the largest possible integer that could appear in the list?
(Note: The median of a set of five positive integers is the middle integer when the
set is arranged in increasing order.)
(A) 15
(B) 16
(C) 17
(D) 24
(E) 25

The correct option is;

(B) 16

Step-by-step explanation:

There are 5 integers with the number 8 occurring more than once

The median = 9

The mean = 10

When we arrange the integers in increasing order we have

A + B + C + D + E

There are at lest two 8s,  with the median = 9

Therefore, we have

A, 8,  8, D, E

The average is 10, and since we have 5 positive integers, therefore

Average = ∑x/n

Where:

n = Count of numbers = 5

∑x = Sum of the 5 numbers

∑x/5 = 10

∴ ∑x = 50

or A + 8 + 8 + D + E = 50

or 8 + 8 + 9 + D + E = 50

D + E = 25

or 8 + 8 + 9 + 9 + 16 = 50 since the median can be gotten at the middle

Therefore, the largest possible integer that could appear on the list = 16.

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.