P=2L+2H
First, isolate 2H:
2H=P-2L
Second, get a singular value of H (just H):
2H/2=(P-2L)/2
Final answer:
H=
First, isolate 2H:
2H=P-2L
Second, get a singular value of H (just H):
2H/2=(P-2L)/2
Final answer:
H=
According to the MLA style, the ______ is a list of sources that are referred directly in a research paper.
2 answers:
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Answer:
The 't' test statistic = 1.46 < 1.69
The test of hypothesis is H 0 :μ = 800 is accepted
A sample of 30 bulbs are found came from average µ= 800
Step-by-step explanation:
Step 1:-
Given population of mean μ = 800
given size of small sample n =30
sample standard deviation 'S' = 45
Mean value of the sample χ = 788
Null hypothesis µ =800
alternative hypothesis µ ≠ 800
<u>Step 2</u>:-
The 't' test statistic t =
t = = 1.4607
<u>Step 3</u>:-
The degrees of freedom γ = n-1 = 30-1 =29
From "t" value from table at 0.05 level of significance ( t = 1.69)
The calculated value t = 1.4607 < 1.69
Therefore The null hypothesis H_{0} ' is accepted.
<u>conclusion</u>:-
A sample of 30 bulbs are came found from average µ= 800
Answer:
The statements about arcs and angles that are true in the figure are;
1) ∠EFD ≅ ∠EGD
2)
3) mFD = 120°
Step-by-step explanation:
1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)
∠CEG = ∠CDG = 90° (Given)
∠GDE = 60° (Given)
∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)
∠ECD = 360° - (90° + 90° + 60°) = 120°
∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)
120° = 2 × ∠EFD
∠EFD = 120°/2 = 60°
∠EFD ≅ ∠EGD
∠ECD = 120°
∠EGD = 60°
∴∠EGD ≠ ∠ECD
2) Given that arc mEF ≅ arc mFD
Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF
∠ECF = mEF = mFD = ∠DCF (Given)
∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)
(Corresponding Parts of Congruent Triangles are Congruent, CPCTC)
∠FED ≅ ∠FDE (base angles of isosceles triangle)
∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)
∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°
∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)
2 × ∠FED = 120°
∠FED = 120°/2 = 60° = ∠FDE
∴ ∠FED = ∠FDE = ∠EFD = 60°
ΔEFD is an equilateral triangle as all interior angles are equal
(definition of equilateral triangle)
3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)
Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)
60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)
60° = 2 × ∠CFE
∠CFE =60°/2 = 30° = ∠CFD
(radii of the same circle)
ΔFCD is an isosceles triangle (definition)
∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)
∠CFD + ∠CDF + ∠DCF = 180°
∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°
mFD = ∠DCF (definition)
mFD = 120°.