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JulsSmile [24]
3 years ago
13

A privately owned lake contains two types of game fish, bass and trout. The owner provides two types of food, A and B, for these

fish. Bass require 2 units of food A and 4 units of food B, and trout require 5 units of food A and 2 units of food B. If the owner has 1200 units of each food, find the maximum number of fish that the lake can support.
Mathematics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer: the maximum number of fish that the lake can support = 375 fish

Step-by-step explanation:

Let b and t represent the bass and trout fish respectively.

Given that;

i. Bass require 2 units of food A and 4 units of food B

ii. trout require 5 units of food A and 2 units of food B

iii. the owner has 1200 units of each food

For food A

2b + 5t = 1200 ....eqn1

For food B

4b + 2t = 1200 ....eqn2

Solving the simultaneous equation

Multiplying eqn1 by 2. We have;

4b + 10t = 2400 .....eqn3

Subtract eqn2 from eqn3

4b-4b +10t-2t = 2400-1200

8t = 1200

t = 1200/8

t = 150

Substituting t= 150 into eqn1

2b + 5(150) = 1200

2b = 1200 - 750 = 450

b = 450/2 = 225

Therefore, the total number of fish is given as

t + b = 150 + 225 = 375

225 bass and 150 trout

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Katyanochek1 [597]

Answer:

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

p_v =P(t_{(5)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation  

The mean and sample deviation can be calculated from the following formulas:

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

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s=1.365 represent the sample standard deviation  

n=6 sample size  

\mu_o =30 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :  

Null hypothesis:\mu \geq 30  

Alternative hypothesis:\mu < 30  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17  

P-value  

We need to calculate the degrees of freedom first given by:  

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Since is a one-side left tailed test the p value would given by:  

p_v =P(t_{(5)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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