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Nikitich [7]
3 years ago
6

In the year 2005, a total of 750 fish were introduced into a manmade lake. The fish population was expected to grow at a rate of

8.3% each year. What will the fish population be in 2050? Round to the nearest whole number. Enter your answer in the box.please help!!!
Mathematics
2 answers:
lyudmila [28]3 years ago
7 0
(100+8.3%=108.3\100=1.083)
difference in years=45
=750 x 1.083 raised exponent45
=27123.25528
rewona [7]3 years ago
6 0
Exponential change can be modeled by:
N = N(0) x λⁿ; where λ is the rate of change. It is greater than 1 when there is growth and less than 1 when there is decay. n is the number of time periods passed.
λ is 1 + 0.083 = 1.083
N(0) is the initial value of 750
n is 2050 - 2005 = 45 years
N = 750 x (1.083)⁴⁵
= 27,000 fish
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Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

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Answer:

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Answer:

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