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Ira Lisetskai [31]
2 years ago
5

A worker is tiling the floor of a rectangular room that is 12 feet by 15 feet. The tiles are square with side lengths 1 & 1/

3 feet. How many tiles are needed to cover the entire floor? Show your reasoning.
Mathematics
2 answers:
Nezavi [6.7K]2 years ago
4 0

Answer:

12/15

Step-by-step explanation:

12/1x1/15=12/15

vfiekz [6]2 years ago
3 0
Gerald Rodgers they are the same
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A rectangle hale card has a length of 3/10 and a width of 4/5 find the perimeter
postnew [5]

Answer:

22/10 or 2 and 1/5

Step-by-step explanation:

To find the perimeter, add all sides together. Since it's a rectangle, the other sides are the same measurement as the ones given. Add 3/10, 3/10, 4/5, and 4/5 together. To do this, find a common denominator. The lowest common denominator is 10 so multiply the 4/5 by 2/2 to get the denominator to be 10. Now, add 3/10, 3/10, 8/10, and 8/10. It adds up to 22/10 but if you don't want an improper fraction, simplify it to 2 and 1/5.

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3 years ago
Look at the graph below.
romanna [79]
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3 years ago
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At the Italian deli, the sandwich maker cut 4 2/3 lb of turkey and 9 3/5 of roast beef. How many more pounds of roast beef were
kherson [118]

Answer:

4 14/15 more pounds of roast beef

Step-by-step explanation:

subtract 4 2/3 from 9 3/5

9 3/5 - 4 2/3= 4 14/15

So your answer is 4 14/15

4 0
3 years ago
A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
igor_vitrenko [27]

Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

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

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, 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)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

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

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

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Which is another name for 23 ten thousands?
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Another name is 230,000
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