For this case we have that by definition, the perimeter of the quadrilateral shown is given by the sum of its sides:
Let "p" be the perimeter of the quadrilateral, then:
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So, the perimeter of the figure is: 
Answer:

Add 2 more parts of yellow
7=2
7=2
7=2
7=2
7=2
---------
35=10
Answer:
Step-by-step explanation:
<u>a)</u>
- Given that ; X ~ N ( µ = 65 , σ = 4 )
From application of normal distribution ;
- Z = ( X - µ ) / σ, Z = ( 64 - 65 ) / 4, Z = -0.25
- Z = ( 66 - 65 ) / 4, Z = 0.25
Hence, P ( -0.25 < Z < 0.25 ) = P ( 64 < X < 66 ) = P ( Z < 0.25 ) - P ( Z < -0.25 ) P ( 64 < X < 66 ) = 0.5987 - 0.4013
- P ( 64 < X < 66 ) = 0.1974
b) X ~ N ( µ = 65 , σ = 4 )
From normal distribution application ;
- Z = ( X - µ ) / ( σ / √(n)), plugging in the values,
- Z = ( 64 - 65 ) / ( 4 / √(12)) = Z = -0.866
- Z = ( 66 - 65 ) / ( 4 / √(12)) = Z = 0.866
P ( -0.87 < Z < 0.87 )
- P ( 64 < X < 66 ) = P ( Z < 0.87 ) - P ( Z < -0.87 )
- P ( 64 < X < 66 ) = 0.8068 - 0.1932
- P ( 64 < X < 66 ) = 0.6135
c) From the values gotten for (a) and (b), it is indicative that the probability in part (b) is much higher because the standard deviation is smaller for the x distribution.
D. Use the formula V=πr^2 h.