Answer:
50
Step-by-step explanation:
Already done this
1. C
2. A
3. B
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<u><em>(Brainliest will be appreciated)</em></u>
Answer: B. 135/364
In solving this kind of probability problems, we have to use the Hyper geometric Probability without replacement. We are given with two groups which would be the source of the 4 members of the new group, 2 coming from each respective group.
Equation:
P = [(pCr)(qCm)]/(sCt)
= [(10C2)(6C2)]/(16C4)
= 135/364
a=(4,1) b=(4,3) c=(8,2) d=(8,0) i think i havent done these in a while
Well we know that the perimeters are equal so:
Psquare = Prectangle
we know perimeter for square is the 4 sides of same lengths added up or 4*s
the perimeter of a rectangle is the then4 sides with 2 being equal added up so 2*L + 2*W
we know that a square side equAls s and it says that s=8x. So:
Psqu = 4 * (8x) (substitute 8x for s)
we know that the rectangle L =10x + 8 and W =10. So
Prect = 2 *(10x + 8) + 2 * (10)
we know :
Prect = Psqu so:
4 * (8x) =2 *(10x + 8) + 2 * (10)
32x = 20x + 16 + 20
12x = 36
x = 3
So if x equals 3 we can sub x in for square perimeter
Psqu = 4 *(8 * 3)
P = 4 * 24
P = 96
we can check this with rectangle perimeter
P = 2 *(10 * 3 + 8) + 2 *10
P = 2 * 38 + 20
P = 96
