For the given cost function C(x), find the oblique asymptote of the average cost function C(X), C(x)=14,000+94x+0.03x^2
1 answer:
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Work the information to set inequalities that represent each condition or restriction.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
<span />
4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.
2) Name the variables.
c: number of color copies
b: number of black-and-white copies
3) Model each restriction:
i) <span>It takes 3 minutes to print a color copy and 1 minute to print a black-and-white copy.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He needs to print at least 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) And must have the copies completed in no more than 12 minutes ⇒</span>
3c + b ≤ 12
<span />
4) Additional restrictions are c ≥ 0, and b ≥ 0 (i.e. only positive values for the number of each kind of copies are acceptable)
5) This is how you graph that:
i) 3c + b ≤ 12: draw the line 3c + b = 12 and shade the region up and to the right of the line.
ii) c + b ≥ 6: draw the line c + b = 6 and shade the region down and to the left of the line.
iii) since c ≥ 0 and b ≥ 0, the region is in the first quadrant.
iv) The final region is the intersection of the above mentioned shaded regions.
v) You can see such graph in the attached figure.
Answer:
A.the type 1 error probability is
B. β = 0.0122
C. β = 0.0000
Step-by-step explanation:
Given that:
Mean = 100
standard deviation = 2
sample size = 9
The null and the alternative hypothesis can be computed as follows:
A. If the acceptance region is defined as , find the type I error probability
.
Assuming the critical region lies within or
, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is
∴
when
From the standard normal distribution tables
Thus, the type 1 error probability is
B. Find beta for the case where the true mean heat evolved is 103.
The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis
Thus;
β = P( type II error) - P( fail to reject )
∴
Given that
From standard normal distribution table
β = 0.0122 - 0.0000
β = 0.0122
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Given that
From standard normal distribution table
β = 0.0000 - 0.0000
β = 0.0000
The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.
2(22 + w) 120
22 + w 60
w 38