The coordinates of the center of this circle will be the average between the extreme points
C = ( (xo+x1)/2 , (yo+y1)/2)
Where,
xo = 3 and x1 = 9
yo = 8 and y1 = 16
Then us stay with"
C = ( (3+9)/2 , (8+16)/2)
C = ( 12/2 , 24/2)
C = ( 6 , 12 )
Using the formula
A = Pe^kt
Where A is the ending amount of population, P is the population k is a constant and 't' is time.
A = 0.39, t= 12 and p= 0.29
0.39= 0.29e^12k
1.34 = e^12k
ln1.34 = 12k
ln1.34/12 = k
K= 0.02
No the graph does not support his claim as k should be equal to 0.5
Answer:
23 chalkboards
Step-by-step explanation:
Given:
Mean length = 5 m
Standard deviation = 0.01
Number of units ordered = 1000
Now,
The z factor =
or
The z factor =
or
Z = - 2
Now, the Probability P( length < 4.98 )
Also, From z table the p-value = 0.0228
therefore,
P( length < 4.98 ) = 0.0228
Hence, out of 1000 chalkboard ordered (0.0228 × 1000) = 23 chalkboard are likely to have lengths of under 4.98 m.