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den301095 [7]
3 years ago
8

Help us neededdddddddd!!

Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Total money out is

27(1.25) + 5(1.99) + 1(1.50) = $45.20

Split 26 ways that's about

45.20/26 = $1.74 each

Of the choices let's choose

Answer: C


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It is defined as the difference between the largest and smallest values in the middle 50% of a set of data<span>. To compute an </span>interquartile range<span> using this definition, first remove observations from the lower quartile. Then, remove observations from the upper quartile.</span>
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3 years ago
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The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4
Mrac [35]

Answer:

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

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Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37

The deviations will be added then.Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Mean = x2 = \frac{41+50+58+62+72+83}{6}

x2 = \frac{366}{6}

x2 = 61

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22

The deviations will be added then.Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

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3.That BC are congruent and that AB and AC are congruent as well

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Let
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