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ratelena [41]
2 years ago
7

The formula P = 0.68x2 - 0.048x + 1 models the approximate population P, in thousands, for a species of fish in a local pond, x

years after 1997. During what year will the population reach 33.984 thousand fish?
Mathematics
1 answer:
marshall27 [118]2 years ago
8 0

Answer:

The answer to your question is 2184

Step-by-step explanation:

Data

Equation    P(x) = 0.68x² - 0.048x + 1

x = years

population = 33984

Process

1.- Substitute the population in the equation

                   33984 = 0.84x² - 0.048x + 1

2.- Equal to zero

                   0.84x² - 0.048x + 1 - 33984 = 0

3.- Simplify

                    0.84x² - 0.048x - 33983 = 0

4.- Solve for x

x = \frac{0.048 +- \sqrt{(0.048^{2}) - (4)(0.84)(-33982)}}{2(0.84)}

x = \frac{0.048 +- \sqrt{337.9}}{1.68}

x1 = 201.2

x2 = -201.2

Only x1 is correct because time can not be negative

5.- Calculate the year

Year = 1997 + 201

Year = 2184

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Step-by-step explanation:

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2 years ago
Segment YB is x+3 units long and segment BW is 2x-9 units long. The diagonal YW is ___ units long.
Gnom [1K]

Answer:

3x - 6

Step-by-step explanation:

Segment YB = x + 3

Segment BW = 2x - 9

Add the two segments.

x + 3 + 2x - 9

3x - 6

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BigorU [14]

The <em><u>correct answer</u></em> is:

She can determine that while there is an association between the variables, there is no correlation, and she cannot determine causation.

Explanation:

Since there is an arch shape to her graph, we know that as one variable changes, the other changes in the same manner. This means there is an association between the variables.

However, since the graph is not linear, there is no correlation between the variables. Since there is no correlation, we cannot determine causation.

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Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model
Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

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Hence, the value of the test statistic is 0.3796.

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Hey there,

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Lastly we divide both sides by 2:

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So the value of x would have to be 5.  

Hope I helped,

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