The integers are 81, 83, 85 and 87.
86.90909090 recurring or 86 and one eleventh.
There are four consecutive odd integers: 81, 83, 85 and 87.
Assuming that series is 88, 87, 85, 81, 73, 57 the next number is 25. The difference each time is twice the previous difference: 88 → 87 : 88 - 87 = 1 87 → 85 : 87 - 85 = 2 = 1 x 2 85 → 81 : 85 - 81 = 4 = 2 x 2 81 → 73 : 81 - 73 = 8 = 4 x 2 73 → 57 : 73 - 75 = 16 = 8 x 2 → next difference is 16 x 2 = 32 → next number is 57 - 32 = 25
.8 .81 .82 .83 .84 .85 .86 .87 .88 .89 .9 There are many more numbers, even an INFINITE amount, between these two.
That's an infinite list. 81, 83, 85, 87, 89, 91 and so on.
The median is 87.
The median is the average of 85 and 87, that is 86.The median is the average of 85 and 87, that is 86.The median is the average of 85 and 87, that is 86.The median is the average of 85 and 87, that is 86.
The integers are 81, 83, 85 and 87.
.81 .82 .83 .84 .85 .86 .87 .89
86.90909090 recurring or 86 and one eleventh.
There are four consecutive odd integers: 81, 83, 85 and 87.
they have won 8 times 81 82 83 85 87 88 89 90
meanAdd up the numbers and divide by how many there are: 87+75+89+81+90+73+85+96+65+92+78+79+83+85+79+90+86+72+74+88+81+85 = 1813There are 22 numbers.⇒ mean = 1813 ÷ 22= 829/22≈ 82.41medianSort the numbers into order; the median is the middle number (if an odd number of numbers) or the mean of the middle two (if an even number of numbers): Sorted: 65, 72, 73, 74, 75, 78, 79, 79, 81, 81, 83, 85, 85, 85, 86, 87, 88, 89, 90, 90, 92, 96There are 22, so the median is the mean of the middle two which are the 11th (83) and 12th (85) numbers:⇒ median = (83+85) ÷2= 84modeThe number which appears the most often. Having sorted the list for the median, it can be seen that the one with the highest frequency appears three times: ⇒ mode = 85
The average of 79, 84, 82, and 87 is 83.
83, 84, 85, 86, 87, 88
only 83 and 89 81 = 9 x 9 85 = 5 x 17 87 = 3 x 29