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3 years ago

Find the value of x.........

Mathematics

2 answers:

Elan Coil [88]3 years ago

4 0

The left triangle:
75 + 50 + x = 180
125 + x = 180
Subtract 125 from both sides.
x = 55 degrees

The center angle on the right triangle is also 55 degrees because the angles are vertical angles.

55 + 85 + x = 180
140 + x = 180
Subtract 140 from both sides.
x = 40 degrees

GREYUIT [131]3 years ago

3 0

The angles of a triangle should add up to 180 degrees. in the left triangle, you should find the 3rd angle

75 + 50 + y = 180
125 + y = 180
y = 55

vertical angles are congruent, so the left angle on the right triangle will also be 55 degrees.

then you do the same thing to find x.

85 + 55 + x = 180
140 + x = 180
x = 40

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4 years ago

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If y=kx, where k is a constant, and y=24 when x=6, what is the value of y when x=5?A. 6B. 15C. 20D. 23

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1 year ago

Over a nine-month period, OutdoorGearLab tested hardshell jackets designed for ice climbing, mountaineering, and backpacking. Ba

Yuri [45]

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(a) The mean is 65.9, the median is, 66.5 and the mode is. 61.

(b) The first and third quartiles are 61 and 71 respectively.

Step-by-step explanation:

The ratings of 20 top-of-the line jackets, arranged in ascending order are as follows:

S = {42 , 53 , 54 , 61 , 61 , 61 , 62 , 63 , 64 , 66 , 67 , 67 , 68 , 69 , 71 , 71 , 76 , 78 , 81 , 83}

(a)

Compute the mean as follows:

$\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}=\frac{1}{20}\times 1318=65.9$

Compute the median as follows:

For an even number of values the median is the average of the middle two values.

$M=\frac{10^{th}obs.+11^{th}obs.}{2}=\frac{66+67}{2}=66.5$

Compute the mode as follows:

The mode of a dataset is the term that occurs most of the time.

$m=61$

Thus, the mean is 65.9, the median is, 66.5 and the mode is. 61.

(b)

The first quartile is the median of the first half of the data set.

S₁ = {42 , 53 , 54 , 61 , 61 , 61 , 62 , 63 , 64 , 66}

$Q_{1}=\text{Median of }S_{1}=\frac{5^{th}obs.+6^{th}obs.}{2}=\frac{61+61}{2}=61$

The third quartile is the median of the second half of the data set.

S₂ = {67 , 67 , 68 , 69 , 71 , 71 , 76 , 78 , 81 , 83}

$Q_{3}=\text{Median of }S_{2}=\frac{15^{th}obs.+16^{th}obs.}{2}=\frac{71+71}{2}=71$

Thus, the first and third quartiles are 61 and 71 respectively.

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The <em>p</em>th percentile is a data value such that at least p% of the data set is less-than or equal to this data value and at least (100 - p)% of the data-set are more-than or equal to this data value.

Use the Excel formula "=PERCENTILE.INC(array,0.9)" to compute the 90th percentile.

The value of the 90th percentile is, 78.3.

This value implies that 90% of the ratings are less than 78.3.

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