Answer:
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 9 - 1 = 8
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.8595
The margin of error is:
M = T*s = 1.8595*2.7 = 5
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 25.3 - 5 = 20.3 milligrams
The upper end of the interval is the sample mean added to M. So it is 25.3 + 5 = 30.3 milligrams.
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.
1 is 47 and 2 is 27 you welcome
Answer:
Please read down below, thanks
Step-by-step explanation:
Vertical angles are always equal.
Because the two expression we have are equal, we can make the equation:
4x + 2 = 5x - 13
Let's move x to one side and the constants to the other side.
Subtract both sides by 4x and add both sides by 13
15 = x
Now we know x = 15, we can solve for the angles.
∠ABC is 4x + 2 so: 4(15) + 2 = 47 degrees
∠CBE added up with ∠ABC is 180 degrees so: 180 - 47 = 133 degrees
∠DBE is the same as ∠ABC so: 47 degrees
∠ABD is also the same as ∠CBE so: 133 degrees