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Arisa [49]
2 years ago
9

A room has a rectangular floor that is 15 feet by 21 feet. What is the area of the floor in square yards ? ​

Mathematics
2 answers:
Soloha48 [4]2 years ago
8 0

Answer:

Hey There!

Let's solve...

Here it is a rectangular floor with length =21 cm and width =15cm

So

area = l \times w \\  = 21 \times 15 \\  = 315 {ft}^{2} \\  \\

So now We need to convert it to yards square

315ft^{2} \times  \frac{ {1yd}^{2} }{ {9ft}^{2} } \\  \\  =  \cancel{315ft^{2}}  \: ^{35}  \times  \frac{ {1yd}^{2} }{ \cancel{9 {ft}^{2} }}  \\  \\  = 35 {yd}^{2}

So how this 9 ft^2 came??

Let's know

ft \to \: yards \\  \\ 3ft = 1yd \\  \\  {3ft}^{2} =  {1yd}^{2}  \\  \\  \boxed{ {9ft}^{2} =  {1yd}^{2}}

<h2>I hope it is helpful to you...</h2><h3>Cheers!_______</h3>
Helen [10]2 years ago
7 0

Answer:

\color{black}{ \rule{500pt}{98888pt}}

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It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele
kow [346]

Answer:

t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

Step-by-step explanation:

1) Data given and notation      

\bar X=23500 represent the sample mean  

s=3900 represent the sample standard deviation      

n=100 sample size      

\mu_o =2000 represent the value that we want to test    

\alpha 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 determine if the true mean is higher than 20000, the system of hypothesis would be:      

Null hypothesis:\mu \leq 2000      

Alternative hypothesis:\mu > 2000      

We don't know the population deviation, so for this case 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{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974      

Calculate the P-value      

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

df=n-1=100-1=99  

Since is a one-side upper test the p value would be:      

p_v =P(t_{99}>8.974)=9.43x10^{-15}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.      

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