To find where on the hill the canonball lands
So 0.15x = 2 + 0.12x - 0.002x^2
Taking the LHS expression to the right and rearranging we have -0.002x^2 + 0.12x -
0.15x + 2 = 0.
So we have -0.002x^2 - 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x^2 + 0.03x -2 = 0.
This is a quadratic eqn with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25. The second solution y = 0.15 * 25 = 3.75
I hope this helps you
-15+6= -9
-9+6= -3
-3+6=3
add 6
Answer:
1001−7
=1001−7
=1001+−7
=994
Step-by-step explanation:=994
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.