(A) 28 . Thanks for the question hope this helps
Answer:
a) 81.5%
b) 95%
c) 75%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 266 days
Standard Deviation, σ = 15 days
We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.
Formula:

a) P(between 236 and 281 days)

b) a) P(last between 236 and 296)

c) If the data is not normally distributed.
Then, according to Chebyshev's theorem, at least
data lies within k standard deviation of mean.
For k = 2

Atleast 75% of data lies within two standard deviation for a non normal data.
Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.
Answer:
let x adult and y student attended
x+y= 94
x=94-y
again
4x+2y = 294 53 adults and 41 student attended
376-4y+2y =294 the event
2y =82
y=41
substituting the value of y in
x+y =94
x= 94-41 = 53
Step-by-step explanation:
Answer:
C. 70°
Step-by-step explanation:
The inscribed angle marked 15° intercepts an arc that is double that measure, so the intercepted arc on the right is 2×15° = 30°.
The external angle marked 20° is half the difference of the intercepted arcs, so is ...
20° = (1/2)(x - 30°)
40° = x - 30° . . . . . . multiply by 2
70° = x . . . . . . . . . . . add 30°
The value of x is 70°.