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KiRa [710]
3 years ago
10

A farmer has 120 feet of fencing available to build a rectangular pen for her pygmy goats. She wants to give them as much room a

s possible to run. Write an expression in terms of a single variable that would represent the area of a rectangle in this family. What are the dimensions of the rectangular pen with the largest area? What is another name for this kind of rectangle?
Mathematics
1 answer:
NemiM [27]3 years ago
7 0

Answer:

It will be a square with dimensions 30 feet by 30 feet.

Step-by-step explanation:

A farmer has 120 feet of fencing available to build a rectangular pen for her pygmy goats.

Let the length of the rectangular pen be = L

Let the width of the rectangular pen be = W

We know the perimeter = 2L+2W

So, we get;

120=2L+2W

A =WL

In terms of single variable we can write this as:

L=(120 -2W)/2

A=W(120-2H)/2

Taking the derivative, dA/dW =60-2W

Setting it to zero to find the critical points ;

60-2W=0

2W=60

W = 30

And 2L+2W=120

2L+2(30)=120

2L+60=120

2L=60

L = 30

So, we get a square with dimensions 30 feet by 30 feet.

And maximum area will be30\times30=900 square feet.

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The mom ate a cup and a half of food


Step-by-step explanation:


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3 years ago
Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

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

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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I Need Help With this FLVS problem can anyone help me and my DBA is tomorrow
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A snail travels 12 centimeters in 4 minutes. The snail travels the same distance each minute. How many milliliters did the snail
Dmitry_Shevchenko [17]

Answer:

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Step-by-step explanation:

Well 12 cenitmters divided by 4 minutes equals the amount of centimeters the snail travels per minute, which is 3 centimeters (12 / 4 = 3). but be careful cuz we need to convert the 3 cm to mm. 1 cm = 10 mm so 3 cm = 30 mm.

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