A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were​ born, and 342 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective?

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.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

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