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il63 [147K]
3 years ago
5

Mr. Magoo needs to paint on of his bedrooms that is 15ft by 20ft. If he paints only the walls, and each gallon of paint covers 1

00ft2, how many gallons of paint will he need?
Mathematics
2 answers:
trasher [3.6K]3 years ago
8 0

To solve this problem you must apply the proccedure shown below:

1. You have that the total area is obtained by applying the formula for calculate the area of a rectangle and multiply it by the four walls:

At=(4)(15ft)(20ft)\\ A=1,200ft^{2}

2. You know that each gallon of paint covers 100 ft², therefore, you can calculate the amount of gallons needed (which you can call x), if you divide the total area of the walls by the area covered by each gallon:

x=\frac{1,200ft^{2}}{100ft^{2}}=12

The answer is: 12 gallons.

photoshop1234 [79]3 years ago
8 0

Answer:

Step-by-step explanation 15x20=300

300/100=3

the answer is 3

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Answer:

t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974    

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Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

\bar X=80 represent the sample mean

s=20 represent the sample standard deviation

n=40 sample size    

\mu_o =86 represent the value that we want to test  

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

z would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

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Alternative hypothesis:\mu \neq 86    

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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".

3) Calculate the statistic    

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

t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974    

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

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In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

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