Premier Bank is planning to have a new building constructed. They would like the length of the building to be 40 feet longer tha
n its width. To fit their needs, each floor of the building will add 12 feet to the height of the building. The cost of the foundation of the building will be $10 per square foot. The cost for each successive floor of the building will be $35,000. They plan to spend $1,180,000 on construction. They would also like the perimeter of the building to be 3 times the height of the building. Create a system of equations to model the situation above, and use it to determine how many of the solutions are viable.
2 answers:
Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.
- L = W + 40
- H = 12×F
- foundation cost = 10×L×W
- cost of floors = 35000×F
- total cost = foundation cost + cost of floors
- total cost = 1,180,000
- P = 2×(L+W)
- P = 3×H
We can rewrite these into a system of equations involving L, W, F.
Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...
... 3H = 2(L+W)
... 3(12F) = 2((W +40) +W) . . . . substitute for H and L
... 36F = 4W +80 . . . . . . . . . . . simplify
... 9F = W +20 . . . . . . . . . . . . . divide by 4
Now, we can look at the cost equations.
... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L
... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10
The two simplified equations in W and F are ...
- 9F = W +20
- 3500F = 118000 -40W -W²
You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.
There is one (nearly feasible) solution:
... W ≈ 180.788
... F ≈ 22.31
There is no combination of floor dimensions and integral numbers of floors that will match <em>all</em> of the problem requirements.
_____
A solution that comes about as close as possible is ...
- length = 221 ft
- width = 181 ft
- perimeter = 804 ft
- # of floors = 22
- height = 264 ft . . . . . (3×height = 792 ft vs. 804 ft)
- cost = $1,170,010 . . . (vs. 1,180,000)
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Answer:
a) Are dependent since we are mesuring at the same individuals but on different times and with a different method
b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.
c)
So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .
Step-by-step explanation:
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".
Let put some notation
x=value for A , y = value for B
A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392
B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411
(a) Are the two samples paired or independent? Explain your answer.
Are dependent since we are mesuring at the same individuals but on different times and with a different method
(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?
The first step is calculate the difference and we obtain this:
d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19
In order to do the qqplot we can use the following R code:
d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)
qqnorm(d)
And the graph obtained is attached.
If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.
(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem
The system of hypothesis for this case are:
Null hypothesis:
Alternative hypothesis:
The second step is calculate the mean difference
The third step would be calculate the standard deviation for the differences, and we got:
The 4 step is calculate the statistic given by :
The next step is calculate the degrees of freedom given by:
Now we can calculate the p value, since we have a two tailed test the p value is given by:
So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .
Answer:
187.5 gallons
Step-by-step explanation:
Make a proportion
=
3000/16 = 187.5, so multiply 1 by 187.5 to find the gallons used in a year
This gets you 187.5 gallons used in a year