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3 years ago

Guys i need these answers badly so i would appreciate if someone responded Thanks !

Mathematics

1 answer:

nikdorinn [45]3 years ago

3 0

Answer:

Y is the midpoint of XZ, so XY = YZ

20x - 30 = 12x + 10
20x - 12x = 10 + 30
8x = 40
x = 5

Find XZ:

XZ = 2(XY) = 2(20*5 - 30) = 140

The answer is:

x = 5 and XZ = 140

------------------------

B is the midpoint of AC, so AB = BC

4x + 3 = 2x + 15
4x - 2x = 15 - 3
2x = 12
x = 6

Find AC:

AC = 2(AB) = 2(4*6 + 3) = 54

The answer is:

x = 6, AC = 54

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Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

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Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

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

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

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