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Charra [1.4K]
2 years ago
12

You work for a landscaper that has a customer needing to seed an area of land 80 feet by 40 feet in size. The garden center has

5-pound bags of grass seed. Each bag of seed can cover 25 square yards of land. You calculate that the area you need to seed is 3200 square feet. You divide by 25 to find that you need 128 bags to seed the area. Is this correct?
Mathematics
2 answers:
Dovator [93]2 years ago
4 0

Answer:

No, the solution discussed in question is not correct.

Step-by-step explanation:

Length of the garden ,l= 80 feet

Breadth of the garden ,b= 40 feet

Area of the rectangle = l × b

Area of the garden = A

A=l\times b = 80 ft\times 40 ft = 3,200 ft^2

Area covered by 1 bag of seeds of grass = 25 yard^2=225 ft^2

1 yard^2 = 9 ft^2

Bags of seed of grass covering area of 3,200 ft^2:

\frac{3200 ft^2}{225 ft^2}=14.22

14.22 bags of seed of grass will occupy the area covered by the garden.

Andreas93 [3]2 years ago
3 0
You need to convert your 3200 sq ft into yards which is 355.556 sq yards then you divide 355.556 by 25 to get 14.22224 bags of seed needed
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Answer:

The F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

Step-by-step explanation:

There are four treatments in the data given, i.e. k = 4.

Total number of observations, n = 12.

Note: degrees of freedom is denoted as df.

For treatment, the degrees of freedom = k-1 = 4-1 =3 df.

The total degrees of freedom = n-1 = 12-1 = 11 df.

The error in degrees of freedom = df (total) - df(treatment)

The error in degrees of freedom = 11 - 3 = 8 df

At α = 0.05 level,from the F table, the F-statistic with (3 , 8)df is 4.07.

Therefore, the F-statistic used to test the hypothesis that the miles per gallon for each fuel are the same is 4.07.

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Small Sample Mean Problem. The maker of potato chips uses an automated packaging machine to pack its 20-ounce bag of chips. At t
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Answer:

Step-by-step explanation:

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2 years ago
A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st
maxonik [38]

Answer:

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

p_v =2*P(z>1.413)=0.158  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

\bar X=5.0611 represent the sample mean

\sigma=0.2803 represent the population standard deviation for the sample  

n=42 sample size  

\mu_o =5 represent the value that we want to test

\alpha=0.1 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 different from 5, the system of hypothesis would be:  

Null hypothesis:\mu = 5  

Alternative hypothesis:\mu \neq 5  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-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:  

z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413    

P-value

Since is a two sided test the p value would be:  

p_v =2*P(z>1.413)=0.158  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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