<span>You just have to break each number with its value.
30+2+.04+.003</span>
Answer:
12m²
Step-by-step explanation:
For a rectangle, with length L and width W,
the perimeter is given as
Perimeter,
P = (2 x Length) + (2 x Width)
P = 2L + 2W
It is given that the perimeter is 48, hence
48 = 2L + 2W (divide both sides by 2)
24 = L + W
or
L = 24 - W -----> eq 1
Also realize that the Area of a Rectangle is given by
A = L x W -----> eq 2
Substituting eq 1 into eq 2,
A = (24 - W) x W
A = -W² + 24W
Recall that for a quadratic equation y = ax² + bx + c, the maxima or minima is given by y(max) = -b/2a
In this case, b = 24 and a = -1
-b/2a = -24/[ 2(-1) ] = 12
Hence for A to be maximum A(max) = 12m² (Answer)
Since it's a multiple of 24, it has to be a multiple of the factors of 24.
Factors of 24:
2,3,4,6,8,12
You can use some of this knowledge to help create the number.
Since the # needs to be a multiple off 2, the last digit needs to be an 8
All numbers that are multiples of 3 have the property that all of their digits added together have to be a number that is evenly divisible by 3.
so your number will look like:
_ _ _ _ _ 8
so start trying combinations for the other 5 digits that give you a number that is a multiple of 3: 3,6,9,12,15, ect. If you can't find one, then it's impossible
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

well then therefore

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

Answer:
a) 0.0082
b) 0.9987
c) 0.9192
d) 0.5000
e) 1
Step-by-step explanation:
The question is concerned with the mean of a sample.
From the central limit theorem we have the formula:

a) 
The area to the left of z=2.40 is 0.9918
The area to the right of z=2.40 is 1-0.9918=0.0082

b) 
The area to the left of z=3.00 is 0.9987

c) The z-value of 1200 is 0
The area to the left of 0 is 0.5

The area to the left of z=1.40 is 0.9192
The probability that the sample mean is between 1200 and 1214 is

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000
e) 
The area to the left of z=-112.65 is 0.
The area to the right of z=-112.65 is 1-0=1