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2 years ago

2x + 5 Find the val 3x+15

Mathematics

1 answer:

Crank2 years ago

3 0

Answer:

Step 1: Find the x

2x + 5: Solve

$2x + 5 - 5 = 0 - 5$

$2x / 2 = -5 / 2$

$x = -5/2$

3x + 15: Solve

$3x + 15 - 15 = 0 - 15$

$3x / 3 = -15 / 3$

$x = -5$

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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 less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than 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 less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

3 0

2 years ago

Read 2 more answers

Jorge is designing a square poster. the poster has an area of 169in^2.what is the length of each side?

Naily [24]

$Let \: the \: side \: of \: the \: poster \: be \: x\: (x>0) \\ \Rightarrow {x}^{2} = 169\Leftrightarrow x = 13 \\ \Rightarrow B. \: 13$