Complete question:
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.
A) Find the mean and standard deviation for the number of girls in groups of 36 births.
B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.
C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?
Answer:
a) mean = 18
Standard deviation =3
b) low range = 12
High range = 24
c) The result of 33 girls is significantly high. Yes, the method is effective.
Step-by-step explanation:
Given:
p = 0.5
n = 36
a) The mean is the product of n and p
Mean u = np
u = 36 * 0.5 = 18
The standard deviation is the square root of the product of n and p&q.
S.d ó = 


b) To find the range rule of thumb:
• For low range
Low range = u - 2ó
= 18 - (2 * 3)
= 12
• High range
= u + 2ó
= 18 + (2*3)
= 24
c) The result is significantly high, because 33 is greater than 24 girls.
A result of 33 girls would prove the method as effective.
Answer:
r = 6 in
Step-by-step explanation:
=
r²h/3 = 216
r² (18/3) = 6 r² = 216
r² = 216/6 = 36
r = 6
Answer:
see below
Step-by-step explanation:
9/10
To get as a fraction over 100, multiply the top and bottom by 10
9*10 90
------ = ------------
10*10 100
9/10 = 90/100
The decimal form for 9/10 is .9 and for 90/100 is .90
The side lengths are 2 units and 4 units. Do you need work?
An arithmetic sequence starts with one number and you add the common difference to the previous term to get the current term
So...
f(x)=mx+b
m=common difference
b=starting point
f(11)=125=11m+b
-
f(1)=5=1m+b
--------
120=10m
Divide both sides by 10
12=m
Your common difference is 12.