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AlladinOne [14]
4 years ago
12

50 POINTS! HELP! A restaurant manager is planning a new layout for the tables in his store. He will be placing tables of 2, tabl

es of 4, and tables of 6 throughout the dining room. To help him decide how many of each he needs, he decides to keep track of all table requests by customers during a 2-week period. Which measure of central tendency would BEST indicate the typical table size needed?
A) mean
B) median
C) mode
D) range
Mathematics
2 answers:
Zarrin [17]4 years ago
6 0

Answer:

mine said c is the correct answer

Step-by-step explanation:

lina2011 [118]4 years ago
3 0

Answer:

The best possible answer for this would be: D. Mode.

This is because mode measures how many times each number appears.

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What is the ordered pair for 7x-2y=-5
iogann1982 [59]

Answer:

Equation is y=7/2x+5/2

Step-by-step explanation:

7x-2y=-5

-7x

-2y = -7x -5

/-2

y = 7/2+5/2

6 0
4 years ago
(1/4) to the 2nd power
Y_Kistochka [10]
You multiply them together; 1/16.
8 0
4 years ago
Renee spent $15.50 on Amazon today and Alex spent $9.69. How much more did Renee spend than Adam
ExtremeBDS [4]

Answer:

$5.81

Step-by-step explanation:

15.50-9.69=5.81

5 0
3 years ago
Read 2 more answers
Charlie makes $34 an hour and will get a 20% raise starting next week. Choose True or False for each statement.
Soloha48 [4]

Answer:

false

Step-by-step explanation:

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7 0
3 years ago
Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (round your answers to the nea
creativ13 [48]
Given a table below which gives the years in A.D. for an archaeological excavation site using the method of tree ring dating:

\begin{tabular}
{|c|c|c|c|c|c|c|c|c|}
1229&1292&1187&1257&1268&1316&1275& 1317&1275
\end{tabular}

Part A:

The sample mean is given by:

\bar{x}= \frac{1229+1292+1187+1257+1268+1316+1275+ 1317+1275}{9} \\  \\  = \frac{11416}{9} =1268

Therefore, the sample mean is 1268.



Part B:

We calculate the sample standard deviation as follows:

s= \frac{1}{9-1} \left(\sqrt{(1229-1268)^2+(1292-1268)^2+(1187-1268)^2+(1257-1268)^2+(1268-1268)^2+(1316-1268)^2+(1275-1268)^2+(1317-1268)^2+(1275-1268)^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{(-39)^2+24^2+(-81)^2+(-11)^2+0^2+48^2+7^2+49^2+7^2}\right)\\ \\ = \frac{1}{8}\left(\sqrt{1521+576+6561+121+2304+49+2401+49}\right)\\ \\ = \frac{1}{8}\left(\sqrt{13,582}\right)\\ \\ = \frac{1}{8}(116.54)=14.57\approx15



Part C:

The 90% confidence interval is given by:

90\% \ C. \ I.=1268\pm1.65\left(\frac{15}{\sqrt{9}}\right) \\  \\ =1268\pm\left(\frac{15}{3}\right)=1268\pm5 \\  \\ =(1268-5, \ 1268+5)=(1263, \ 1273)
8 0
3 years ago
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