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Blizzard [7]
3 years ago
13

a closet contains 8 pairs of pants 10 shirts and 4 pair of shoes. How many leaves on a tree diagram would represent all possible

combination
Mathematics
2 answers:
poizon [28]3 years ago
5 0
The answer is, there are 320 possible combinations of outfits
Anna007 [38]3 years ago
5 0

Answer: Hello mate!

you have:

8 pairs of pants, 10 shirts and 4 pairs of shoes:

We want to find the total number of possible combinations, where you use a pair of pants, a shirt and a pair of shoes.

These combinations are equal to the product between the number of options for each event (where the events are "selectinc a pair of pants", "selecting a shirt", etc)

For pants you have 8 options, for shirts 10 options and for shoes 4 options, then the total number of combinations are:

8*10*4 = 8*40 = 320

Then there are 320 possible combinations, in a tree diagram you have a leave for every combination, so you have 320 leaves.

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The area of the figure is 30 because u have to add an subtract the numbers on the side
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Shards of clay vessels were put together to reconstruct rim diameters of the original ceramic vessels found at the Wind Mountain
Otrada [13]

Answer:

a) In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n-1}} (3)  

The mean calculated for this case is

The sample deviation calculated s=3.46 \approx 3.5

b) 15.8-1.383\frac{3.5}{\sqrt{10}}=14.27    

15.8+1.383\frac{3.5}{\sqrt{10}}=17.33    

So on this case the 80% confidence interval would be given by (14.27;17.33)  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)^2}{n-1}} (3)  

The mean calculated for this case is

The sample deviation calculated s=3.46 \approx 3.5

Part b

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.80 or 80%, the value of \alpha=0.2 and \alpha/2 =0.1, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,9)".And we see that t_{\alpha/2}=1.383

Now we have everything in order to replace into formula (1):

15.8-1.383\frac{3.5}{\sqrt{10}}=14.27    

15.8+1.383\frac{3.5}{\sqrt{10}}=17.33    

So on this case the 80% confidence interval would be given by (14.27;17.33)    

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Step-by-step explanation:


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