Answer:
The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the student's t-distribution to find the confidence interval.
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 = 16 - 1 = 15
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 35 degrees of freedom(y-axis) and a confidence level of ). So we have T = 2.9467
The margin of error is:
M = T*s = 2.9467*0.058 = 0.171
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 1.053 - 0.171 = 0.882kg
The upper end of the interval is the sample mean added to M. So it is 1.053 + 0.171 = 1.224 kg.
The 99% two-sided confidence interval for the average sugar packet weight is between 0.882 kg and 1.224 kg.
Answer:
Approximately 6.4
Step-by-step explanation:
We can use the pythagorean thereom here, that tells us (a^2)+(b^2)=c^2. C is the hypotenuse, the side opposite from the right angle, while a and b are the other sides. We can insert 5 and 4 as a and b, and solve for c
:(5^2)+(4^2)=c^2
:25+16=c^2
:41=c^2
:sqrt(41)=6.4=c (We square rooted both sides. 6.4 is only rounded to the nearest hundredths place.) Hope this helps!
Answer:
6.24
Step-by-step explanation:
A= 18.42g, B = 5.8g,C = 0.75g
Total = 18.42 + 5.8 + 0.75
= 24.97g
Then divided into four equal parts
= 24.97g/4
= 6.2425g
= 6.24g
Answer:
4 times
Step-by-step explanation:
You divide 12 by 3