Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):

since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:

Now plug in p and a values:


therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10
D) 5 feet, and 10 feet
Answer:
m=6
Step-by-step explanation:
For this figure to be a parallelogram, the sides must be the same length
4m+2 = 3m+8
Subtract 3m from each side
4m+2-3m = 3m+8-3m
m+2 =8
Subtract 2 from each side
m+2-2 =8-2
m=6
So the equation of a circle is (x - h)² + (y - k)² = r² where (h,k) are the coordinates of the center of the circle and r is the radius. The diameter of a circle is a line that goes from one point of the circle to the other through the center of the circle. Well the center would be midway through the diameter so use midpoint formula to find the center which is (h,k) Mid point formula is both given x's added together divided by 2 for h and both y coordinates added together divided by 2 to find k
(10+0)/2
10/2= 5
(12+2)/2
14/2 = 7
so the center of the circle is (5,7) now use distance formula using the center and one of the points to the radius
√((5-10)²+(7-12)²)
√(-5²+ -5²)
√(25 + 25)
√50 is the radius
Now plug all found information into circle equation
(x-5)² + (y-7)² =50 note the end is 50 because the circle equation is radius squared and since the radius is √50, radius² is 50.
Answer is c
ANSWER

EXPLANATION
The given expression is

We multiply out to obtain,

Recall that,

We apply this rule to obtain,

This simplifies to
Answer:
Mean = 94
Standard deviation = 1.12
The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation, which is also called standard error 
In this problem, we have that:

By the Central Limit Theorem
The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.
Mean = 94
Standard deviation:
