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mina [271]
3 years ago
7

Following the lead of the National Wildlife Federation, the Department of the Interior of a South American country began to reco

rd an index of environmental quality that measured progress and decline in the environmental quality of its forests. The index for the years 2004 through 2014 is approximated by the functionI I(t) = 1/3t^3 − 5/2t^2 + 80 (0 ≤ t ≤ 10) where t = 0 corresponds to 2004. For what years is the quality of the forests in this country improving?
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
7 0

Answer:

After 2009 the quality of the forests in this country improving.

Step-by-step explanation:

Consider the provided function.

f(t) = \frac{1}{3}t^3-\frac{5}{2}t^2 + 80

f'(t) = t^2-5t

Substitute f'(t)=0.

t^2-5t=0

t(t-5)=0

t=0\ or\ t=5

It is given that  (0 ≤ t ≤ 10)

The test interval are (0,5) and (5,10)

For (0 < t < 5) the value of f'(t)

For (5 < t < 10) the value of f'(t) >0

Hence, the function is increasing from (5,10).

t = 0 corresponds to 2004.

Therefore, after 2009 the quality of the forests in this country improving.

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The following table presents the time taken to review articles that were submitted for publication to a particular journal durin
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A) Variance = 10.24

B) Standard Deviation = 3.2

One of the measurements of dispersion is the standard deviation, which is exclusively used for quantitative data. It aids in determining if the data's mean is a suitable measurement to reflect the core value.

TIME         FREQUENCY(f)  MIDPOINT(x)    d    d²     fd              fd²

0 - 0.9         43                          0.45             -3     9     -129            387

1.0 - 1.9         17                           1.45              -2    4     -34                68

2.0 - 2.9       19                            2.45             -1     1       -19               19

3.0 - 3.9        18                            3.45             0     0       0                  0

4.0 - 4.9        14                            4.45              1      1        14               14

5.0 -5.9         16                            5.45              2     4        32             64                            

∑f = 127                                                              

∑fd = -136  

∑fd² = 552

d = \frac{x - 3.45}{1}

Standard Deviation = \sqrt \frac{fd^{2} }{N} - \sqrt \frac{(fd)^{2} }{N} * i

\sqrt \frac{552}{127} -\sqrt (\frac{-136}{127})^{2}   * 1

√4.35 - √1.15

Standard Deviation = 3.2

(SD)² = (3.2)² = 10.24

Variance = 10.24

To know more about standard deviation, refer to this link :

brainly.com/question/12402189

#SPJ1

7 0
1 year ago
Muffins cost $3.00 per dozen.
GenaCL600 [577]

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