Answer:
The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of
.
In the study 380 babies were born, and 342 of them were girls.
This means that
99% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:

As percentages:
0.8604*100% = 86.04%.
0.9396*100% = 93.96%.
The 99% confidence interval estimate of the percentage of girls born is (86.04%, 93.96%). Considering the actual percentage of girls born is close to 50%, the percentage increased considerably with this method, which means that it appears effective.
Answer:
2/25. Hope this helps :) :) :) :)
Answer:
18
Step-by-step explanation:
Hello,
In order to solve for x, we need to isolate the variable on one side of the equation.
To do this, we subtract 450 from both sides of the equation, and this will give you:
28.75x = 517.5
As mentioned, we need to isolate x, so we divide both sides by 28.75. This will give the following.
x = 18
Hope this helps!
Answer:
If a<b, x<2
If a>b, x>2
Step-by-step explanation:
ax-bx>2a-2b, (a-b)x>2(a-b)
If a>b, x>2(a-b)/(a-b)=2
If a<b, x<2(a-b)/(a-b)=2
Answer:
92905
Step-by-step explanation:
27,500=0.296e
29.6%=0.296
27,500/0.296 =e
e rounded to the nearest whole number is 92,905