The set that is more variable is the daily temperature as they won't be clustered but rather spread out.
<h3>How to know the set that's more variable?</h3>
The mean of the temperature will be:
= (51 + 37 + 53 + 27 + 91 + 50 + 50 + 83 + 83 + 90)/10
= 615/10
= 61.5
The mean of the precipitation will be:
= (4.5 + 3.8 + 1.4 + 2.4 + 2.8 + 4.6 + 4.7 + 2.3 + 1.5)/10
= 28/10
= 2.8
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Since we are given the values for both rows only in the second column, we can use this to solve for the rest of the missing values. Simply divide 3 by 2.49 and multiply that resultant by and multiply that by any missing value for cookies to find the missing cost. In order to solve for the cookies when cost is given, divide 2.49 by 3 and multiply that resultant. I will solve for the first, third, and fourth columns.
3/29=1.2
For column 1, 1.2*1=1.20
For column 3, 1.2*20=24.10
For column 4, 1.2*100=12.00
Well, 2 is a whole number so you keep that as a whole number. 3/4 is like money 4 quarters equals a dollar so how much is 3 quarters. So the answer is 2.75
(x+9)(x+8) = x^2 +17x + 72
(x- 7)(x+8) = x^2 + x - 72
(x- 5)(x- 6) = x^2 -11x + 30
(2x + 3)(3x+2) = 6x^2 + 13x +6
(5x - 4)(2x-5) = 10x^2 - 33x + 20
(x-4)^2 = x^2 -8x + 16
(2x+1)^2 = 4x^2 +4x + 1
(4x+3)(4x-3) = 16x^2 - 9
2x2x5x5 or 2^2*5^5 hope this helps!
