See the attached image for the graph. Specifically figure 2 is the graph you want. You can leave the red points on the graph or decide to erase them (leave behind the blue line though).
To generate each of the red points, you'll plug in various x values to get corresponding y values.
For instance, plug in x = 0 and we get...
y = -|x-6| - 6
y = -|0-6| - 6
y = -|-6| - 6
y = -6 - 6
y = -12
So when x = 0, the y value is -12. The x and y value pair up to get (x,y) = (0,-12)
Another example: plug in x = 2
y = -|x-6| - 6
y = -|2-6| - 6
y = -|-4| - 6
y = -4 - 6
y = -10
So the point (2,-10) is on the graph
The idea is to generate as many points as possible so we get an idea of what this thing looks like.
Generate enough points, and you'll get what you see in Figure 1 (see attached image)
Then draw a line through all of the points. The more points you use, the more accurate the drawing. Doing that will generate the blue function curve you see in Figure 2 (also attached)
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:
11:13
Step-by-step explanation:
132:156
Split in half
66:78
Again
33:39
By 3
11:13
1/(x^5) is equal to x^-5, by the definition of negative exponents.
Substitute the given values for the parameters in the point-slope form of the equation of a line. That form is
y - k = m(x - h)
for point (h, k) and slope m.
You have (h, k) = (1, 2) and m=-3, so your equation is
y - 2 = -3(x - 1)