Area = 4x(2x + 3) = 8x2 + 12x
Perimeter = 2(2x + 3 + 4x) = 12x + 6
Answer:
-18
Explanation:
a - b
a = -7
b = 11
Insert -7 for 'a' and 11 for 'b'
-7 - 11
= -18
Answer:
2550yards
Step-by-step explanation:
Perimeter of a rectangle = 2L + 2W
L is the length
W is the width
If perimeter of a rectangle is 425 yards, then;
425 = 2L + 2W
425 = 2(L+W)
L+W = 425/2 .... 1
If the length and width of the rectangle are each multiplied by 6
P = 6(2L) + 2(2W)
P = 12L + 12W
P = 12(L+W) .... 2
Substitute 1 into 2
P = 12(425/2)
P = 6*425
P = 2550yd
Hence the perimeter of new triangle is 2550yards
Answer:

--- Variance
Step-by-step explanation:
Given

Solving (a): Calculate the mean.
The given data is a grouped data. So, first we calculate the class midpoint (x)
For 51 - 58.

For 59 - 66

For 67 - 74

For 75 - 82

For 83 - 90

So, the table becomes:

The mean is then calculated as:



-- approximated
Solving (b): The sample variance:
This is calculated as:

So, we have:


-- approximated
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