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inysia [295]
3 years ago
13

Alexandra wishes to surround his rectangular flower bed with a gravel walk 1 meter wide flower bed is 5m and the width is 4m. Wh

at will be the area of the gravel walk?
Mathematics
1 answer:
Brut [27]3 years ago
8 0
Answer : 8

Explanation: If the area of the rectangle is 5x4 and the gravel walk would be 1 meter on each side you subtract 5-1 and 4-3 to get the 4x3. This is now the inside area or where the flowers are. Now multiply 5x4 (20) to find the total area of the flower bed and 4x3 (12) to find the inside area where the gravel is not. Subtract these (20-12) to find the area of the gravel (8). :)
Hope this helps
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Viefleur [7K]

Answer:

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082    

p_v =2*P(Z>4.082)=4.46x10^{-5}  

So the p value is a very low value and using any significance level for example \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.

Step-by-step explanation:

1) Data given and notation  

X_{1}=30 represent the number of people with a characteristic in 1

X_{2}=10 represent the number of people with a characteristic in 2

n_{1}=50 sample of 1 selected  

n_{2}=50 sample of 2 selected  

p_{1}=\frac{30}{50}=0.6 represent the proportion of people with a characteristic in 1

p_{2}=\frac{10}{50}=0.2 represent the proportion of people with a characteristic in 2

z would represent the statistic (variable of interest)  

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion 1 is different from proportion 2 , the system of hypothesis would be:  

Null hypothesis:p_{1} = p_{2}  

Alternative hypothesis:p_{1} \neq p_{2}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{30+10}{50+50}=0.4  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082    

4) Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test.    

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

p_v =2*P(Z>4.082)=4.46x10^{-5}  

So the p value is a very low value and using any significance level for example \alpha=0.05 always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.  

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