Find the value of x.
1 answer:
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Answer:
P value = 0.1575
Step-by-step explanation:
Data provided in the question
t = 1.021
n = 33
= 98.6° F
Based on the above information, the P-value could be determined by using the excel spreadsheet i.e. shown in the attachment
So, the P value is 0.1575
The 1.021 denotes the test statistic
32 denotes degrees of freedom
And,
1 denotes one tailed alternative hypothesis.
Answer:
∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9
Step-by-step explanation:
From Sine rule
∴ b / sinB = c / sinC
From the question,
b = 129, c = 168 and ∠B = 48°
∴ 129 / sin48° = 168 / sinC
Then, sinC = (168×sin48)/129
sinC = 0.9678
C = sin⁻¹(0.9678)
C = 75.42
∠C2=75.4°
and
∴∠C1 = 180° - 75.4°
∠C1 =104.6°
For ∠A
∠A1 = 180° - (104.6°+48°) [sum of angles in a triangle]
∠A1 = 27.4°
and
∠A2 = 180° - (75.4° + 48°)
∠A2 = 180° - (123.4°)
∠A2 = 56.6°
For side a
a1/sinA1 = b/sinB
a1/ sin27.4° = 129/sin48
a1 = (129×sin27.4°)/sin48
a1 = 79.8845
a1 = 79.9
and
a2/sinA2 = b / sinB
a2/ sin56.6° = 129/sin48
a2 = (129×sin56.6°)/sin48
a2 = 144.9184
a2 = 144.9
Hence,
∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9
Answer:
∠JKL = 38°
Step-by-step explanation:
PQRS, JQK and LRK are straight lines
Let's take the straight lines in the diagrams one after the other to find what they consist.
The related diagram can be found at brainly (question ID: 18713345)
Find attached the diagram used for solving the question.
For straight line PQRS,
2x°+y°+x°+2y° = 180°
(Sum of angles on a Straight line = 180°)
Collect like terms
3x° + 3y° = 180°
Also straight line PQRS = straight line PQR + straight line SRQ
For straight line PQR,
2y + x + ∠RQM = 180° ....equation 1
For straight line SRQ,
2x + y + ∠MRQ = 180° ....equation 2
Straight line PQRS = addition of equation 1 and 2
By collecting like times
3x +3y + ∠RQM + ∠MRQ = 360°....equation 3
Given ∠QMR = 33°
∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)
∠RQM + ∠MRQ + 33° = 180°
∠RQM + ∠MRQ = 180-33
∠RQM + ∠MRQ = 147° ...equation 4
Insert equation 4 in 3
3x° +3y° + 147° = 360°
3x +3y = 360 - 147
3x +3y = 213
3(x+y) = 3(71)
x+y = 71°
∠JQP = ∠RQK = 2y° (vertical angles are equal)
∠LRS = ∠QRK = 2x° (vertical angles are equal)
∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)
2x+2y + ∠QKR = 180
2(x+y) + ∠QKR = 180
2(71) + ∠QKR = 180
142 + ∠QKR = 180
∠QKR = 180 - 142
∠QKR = 38°
∠JKL = ∠QKR = 38°
