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grin007 [14]
3 years ago
10

Ingnore question 17 but i need help 30 POINTS!!!

Mathematics
1 answer:
algol [13]3 years ago
5 0
Not sure why they put the x bit

ok so know this:
AGE=HGB=CHG=FHD
and
AGH=EGB=CHF=GHD
and
AGE+AGH=180


11.
AGE=FHD=30

12.
AGH=CHF=150

13.
CHF=BGE=110

14.
CHG+HGA=180
CHG+120=180
CHG=60

15.
BHG=3x=AGE=CHG=FHD
not sure, but if we did use the fact that BGH+DHG=180 and BGH=3x and DHG=2x+50 then BHG=78 degrees

16.
GHD=2x+50=CHF=EGB=AGH, if we use facts above then GHD=102 degrees
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(a) The mean is 52 and the median is 55.

(b) The first quartile is 44 and the third quartile is 60.

(c) The value of range is 33 and the inter-quartile range is 16.

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(e) There are no outliers in the data set.

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The data provided is:

S = {55, 56, 44, 43, 44, 56, 60, 62, 57, 45, 36, 38, 50, 69, 65}

(a)

Compute the mean of the data as follows:

\bar x=\frac{1}{n}\sum x\\=\frac{1}{15}[55+ 56+ 44+ 43+ 44+ 56+ 60+ 62+ 57+ 45 +36 +38 +50 +69+ 65]\\=\frac{780}{15}\\=52

Thus, the mean is 52.

The median for odd set of values is the computed using the formula:

Median=(\frac{n+1}{2})^{th}\ obs.

Arrange the data set in ascending order as follows:

36, 38, 43, 44, 44, 45, 50, 55, 56, 56, 57, 60, 62, 65, 69

There are 15 values in the set.

Compute the median value as follows:

Median=(\frac{15+1}{2})^{th}\ obs.=(\frac{16}{2})^{th}\ obs.=8^{th}\ observation

The 8th observation is, 55.

Thus, the median is 55.

(b)

The first quartile is the middle value of the upper-half of the data set.

The upper-half of the data set is:

36, 38, 43, 44, 44, 45, 50

The middle value of the data set is 44.

Thus, the first quartile is 44.

The third quartile is the middle value of the lower-half of the data set.

The upper-half of the data set is:

56, 56, 57, 60, 62, 65, 69

The middle value of the data set is 60.

Thus, the third quartile is 60.

(c)

The range of a data set is the difference between the maximum and minimum value.

Maximum = 69

Minimum = 36

Compute the value of Range as follows:

Range =Maximum-Minimum\\=69-36\\=33

Thus, the value of range is 33.

The inter-quartile range is the difference between the first and third quartile value.

Compute the value of IQR as follows:

IQR=Q_{3}-Q_{1}\\=60-44\\=16

Thus, the inter-quartile range is 16.

(d)

Compute the variance of the data set as follows:

s^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{15-1}[(55-52)^{2}+(56-52)^{2}+...+(65-52)^{2}]\\=100.143

Thus, the variance is 100.143.

Compute the value of standard deviation as follows:

s=\sqrt{s^{2}}=\sqrt{100.143}=10.01

Thus, the standard deviation is 10.01.

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An outlier is a data value that is different from the remaining values.

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Q_{1}-1.5QR=44-1.5\times 16=20

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Q_{3}+1.5QR=60-1.5\times 16=80

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None of the values is less than 20 or more than 80.

Thus, there are no outliers in the data set.

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