Find the m of
1 answer:
Both of these problems will be solved in a similar way, but with different numbers. First, we set up an equation with the values given. Then, we solve. Lastly, we plug into the original expressions to solve for the angles.
[23] ABD = 42°, DBC = 35°
(4x - 2) + (3x + 2) = 77°
4x+ 3x + 2 - 2 = 77°
4x+ 3x= 77°
7x= 77°
x= 11°
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ABD = (4x - 2) = (4(11°) - 2) = 44° - 2 = 42°
DBC = (3x + 2) = (3(11°) + 2) = 33° + 2 = 35°
[24] ABD = 62°, DBC = 78°
(4x - 8) + (4x + 8) = 140°
4x + 4x + 8 - 8 = 140°
4x + 4x = 140°
8x = 140°
8x = 140°
x = 17.5°
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ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°
DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°
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Answer: School Y has 4 points and School X has 12 points.
Explanation:
Let the number of points School Y has be x
Let the number of points School X has be 3x
According to question, we get
So, the number of point School Y has 4.
and number of points School X has
Hence, School Y has 4 points and School X has 12 points.
Answer:
(a) The cumulative frequency curve for the data is attached below.
(b) (i) The inter-quartile range is 10.08.
(b) (ii) The 70th percentile class scores is 0.
(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.
Step-by-step explanation:
(a)
To make a cumulative frequency curve for the data first convert the class interval into continuous.
The cumulative frequencies are computed by summing the previous frequencies.
The cumulative frequency curve for the data is attached below.
(b)
(i)
The inter-quartile range is the difference between the third and the first quartile.
Compute the values of Q₁ and Q₃ as follows:
Q₁ is at the position:
The class interval is: 34.5 - 39.5.
The formula of first quartile is:
Here,
l = lower limit of the class consisting value 25 = 34.5
(CF) = cumulative frequency of the previous class = 24
f = frequency of the class interval = 20
h = width = 39.5 - 34.5 = 5
Then the value of first quartile is:
The value of first quartile is 34.75.
Q₃ is at the position:
The class interval is: 44.5 - 49.5.
The formula of third quartile is:
Here,
l = lower limit of the class consisting value 75 = 44.5
(CF) = cumulative frequency of the previous class = 74
f = frequency of the class interval = 15
h = width = 49.5 - 44.5 = 5
Then the value of third quartile is:
The value of third quartile is 44.83.
Then the inter-quartile range is:
Thus, the inter-quartile range is 10.08.
(ii)
The maximum upper limit of the class intervals is 69.5.
That is the maximum percentile class score is 69.5th percentile.
So, the 70th percentile class scores is 0.
(iii)
Compute the probability that a student scored at most 50 on the examination as follows:
Thus, the probability that a student scored at most 50 on the examination is 0.89.
