The strawberry lace is 36 cms long.
36 cms = 360 mm
(Because we multiply by 10, when we convert from cm to mm).
Answer:
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Step-by-step explanation:
Given that :
Mean = 265
standard deviation = 14
The formula for calculating the z score is 
x = μ + σz
At middle of 50% i.e 0.50
The critical value for 
From standard normal table
+ 0.67 or -0.67
So; when z = -0.67
x = μ + σz
x = 265 + 14(-0.67)
x = 265 -9.38
x = 255.62
when z = +0.67
x = μ + σz
x = 265 + 14 (0.67)
x = 265 + 9.38
x = 274.38
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days



now, with that template above in mind, let's see this one

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units
check the picture below
The answer is 3.9 x 10^-5