Let the smallest of 4 consecutive odd numbers be 2 n + 1 , where n is an integer. Show, using algebra, that the sum of any 4 consecutive odd numbers is always a multiple of 8. Give your answer as an expression that makes explicit that it is a multiple of 8.

Expression is 8(n+2)

Step-by-step explanation:

smallest of 4 consecutive odd numbers =2n + 1

consecutive odd integers are found by adding 2 to the any given odd numbers

Thus, 2nd consecutive odd integers  = 2n + 1 + 2 = 2n + 3

3rd consecutive odd integers  = 2n + 3 + 2 = 2n + 5

2nd consecutive odd integers  = 2n + 5 + 2 = 2n + 7

Thus, 4 consecutive odd integers are

2n + 1 ,2n + 3,2n + 5,2n + 7

sum of these numbers are = 2n + 1 +2n + 3 + 2n + 5+2n + 7 = 8n+16

sum of these numbers are = 8(n+2)

Thus, we see that the sum of numbers are 8(n+2)

As, 8 is common for n+2, whatever is value of n, the number will be multiple of 8 .

thus expression is 8(n+2)

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step-by-step explanation:

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.