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SCORPION-xisa [38]
3 years ago
7

You are building a rectangular garden against the back of your house, using 80 feet of fencing. The area of the garden needs to

be greater than 400 square feet but less than 600 square feet.

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
8 0

Answer:

The answer is d, (20 - 10\sqrt{2}, 10) ∪ (30, 20 + 10\sqrt{2})

Step-by-step explanation:

We know that we have 80 feet of fencing available, and that we really only need to have fencing around 3 sides of the garden.

80 feet will equal the sum of the horizontal side length (long one) and the two vertical side lengths. Represent this as 80 = 2x + y, where x is the short length, and y is the length of the long side of fencing. There is a 2x because we need two short sides of fencing and only one long side (y) because we don't need another long side of fencing closest to our house.

We are looking for possible lengths of the long side, y. Manipulate the equation such that y is isolated:

y = -2x + 80

Now the tricky part. We know that the area of the garden must be greater than 400 and less than 600. We can represent this situation with:

400 < xy < 600, where xy is the area (x represents width/short sides, and y represents length/the long side, remember that area = length times width)

This compound inequality is impossible to solve. But, we know that we have already written y in terms of x from before. (y = -2x + 80) Using this, we can substitute y in this compound inequality with this equation. So,

400 < x(-2x + 80) < 600

= 400 < -2x² + 80x < 600

Now, we go about the regular process in solving quadratic inequalities, in this case, we have a compound inequality.

Solving for the left side, we get 0 < -2x² + 80x - 400 (I subtracted 400 from both sides) Keep this inequality in mind.

Solving for the right side, we get -2x² + 80x - 600 < 0, which is the same as 0 > -2x² + 80x - 600. (I subtracted 600 from both sides) Keep this inequality in mind.

You can get crazy with the quadratic equation, but I didn't do that on this quiz, as I might not have even finished on time lol. I just chose to graph both quadratic functions on the Edgenuity graphing calculator, and looked at where each of the two functions had x values that made the inequality true:

where the first function, -2x² + 80x - 400, had positive y values. This starts after x = 5.858 (equivalent to 20 - 10\sqrt{2} from the answer choices lol)

a little bit to the right of that, we see where the second function starts to have positive values, which is after x = 10. but remember, we don't care about when this function has positive y values, we care about when it has negative y values.

this forms the first interval, (20 - 10\sqrt{2}, 10). 10 is there because up until x=10, the inequality is satisfied.

to the right of this, we see that the second function (looks a little more squished) touches the x axis again at 30, and after 30, the function becomes true according to the inequality and starts to have negative values, and will continue to so for positive infinity.  Keep 30 in mind.

and a little to the right of that, we have that the first function (the wider one) touches the x axis at x = 34.142, equivalent to 20 + 10\sqrt{2} from the answer choices(I punched this into a calculator to see) and then begins to have negative values after this, making the inequality not true.

This forms the second interval, (30, 20 + 10\sqrt{2}). That's how I got my answer.

Think back to the Edgenuity lesson on function modeling. We had to find areas on a graph where all functions were satisfied called the feasible region. This problem builds on those skills, and the "feasible regions" if you're looking at the two functions graphed, are the two areas where there is a thin gap between the line of the first and second function. The answer to this problem is the "domains" if you'd like to think of it that way, of these two areas where both inequalities are true.

I hope this helps. Edgenuity precal is getting harder and harder.

Minchanka [31]3 years ago
3 0

Answer:

is d

Step-by-step explanation:

got a 100

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Answer:

t=\frac{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73    

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Data given and notation    

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s=8.6 represent the population standard deviation    

n=45 sample size    

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\alpha=0.01 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)    

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What are H0 and Ha for this study?    

Null hypothesis:  \mu \leq 83.2  

Alternative hypothesis :\mu > 83.2  

Compute the test statistic  

The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{\sigma}{\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{86.7-83.2}{\frac{8.6}{\sqrt{45}}}=2.73    

Give the appropriate conclusion for the test  

First we need to calculate the degrees of freedom given by:

df=n-1=45-1=44

Since is a one side right tailed test the p value would be:    

p_v =P(t_{44}>2.73)=0.0045    

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If we compare the p value and a significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is signficantly higher than 83.2 at 1% of significance.

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