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3 years ago

10

a deli served 1288 sandwiches in 4 weeks. if it served the same number of sandwiches each day,how many sandwiches did it serve i

n 1 day?

1 answer:

pshichka [43]3 years ago

6 0

1288/(4x7)
=1288/28
=46 sandwiches

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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,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

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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