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:
4 cookies
Step-by-step explanation:
Let the number of cookies he sold be c and that of brownies be b.
Assuming that brownies and cookies are the only type of baked goods sold,
b +c= 10 -----(1)
Amount of money received= $20
b(cost of brownie) +c(cost of cookie)= $20
2.50b +1.25c= 20 -----(2)
From (1): b= 10 -c -----(3)
Substitute (3) into (2):
2.50(10 -c) +1.25c= 20
Expand:
2.50(10) +2.50(-c) +1.25c= 20
25 -2.50c +1.25c= 20
-1.25c +25= 20
Being constants to 1 side:
-1.25c= 20 -25
-1.25c= -5
c= -5 ÷(-1.25)
c= 4
Thus, Joe sold 4 cookies.
Use the FOIL method (first, outside, inside, last) to multiply:
2x^2-8x+3x-12
This can be simplified even more into:
2x^2-5x-12.
Hope this helped!