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lys-0071 [83]
2 years ago
7

A model predicts the profit, p of a company as p(x) = x3 – 3x2 – 16x + 48, where p is in thousands of dollars and x is number of

years passed since January 1, 2011. For which time interval would the company make a loss?
Mathematics
1 answer:
zavuch27 [327]2 years ago
3 0
The first step is to determine the zeros of p(x).
From the Remainder Theorem, 
p(a) = 0  => (x-a) is a factor of p(x), and x=a is a zero of p(x).

Try x=3:
p(3) = 3^3 - 3*3^2 - 16*3 + 48 = 27 - 27 - 48 + 48 = 0
Therefore x=3 is a zero, and (x-3) is a factor of p(x).

Perform long division.
                    x²             -  16
      -------------------------------------
x-3 |  x³  -  3x²  -  16x  +  48
         x³  -  3x²
        -----------------------------------
                         -  16x  + 48
                         -  16x  +  48         

Note that x² - 6 = (x+4)(x-4).

Therefore the complete factorization of p(x) is
p(x) = (x-3)(x+4)(x-4)

To determine when p(x) is negative, we shall test between the zeros of p(x)
   x      p(x)       Sign
----  ---------    ---------
  -4          0    
   0        48         +
   3           0
3.5   -1.875         -
   4           0

p(x) is negative in the interval x = (3, 4).

Answer
The time interval is Jan. 1, 2014 to Jan. 1, 2015.
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AysviL [449]

Answer:

The mean calculated for this case is \bar X=584

And the 95% confidence interval is given by:

584-2.776\frac{86.776}{\sqrt{5}}=476.271    

584+2.776\frac{86.776}{\sqrt{5}}=691.729    

So on this case the 95% confidence interval would be given by (476.271;691.729)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=584

The sample deviation calculated s=86.776

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=5-1=4

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

584-2.776\frac{86.776}{\sqrt{5}}=476.271    

584+2.776\frac{86.776}{\sqrt{5}}=691.729    

So on this case the 95% confidence interval would be given by (476.271;691.729)    

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