Find c and round. geometry
2 answers:
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(C) 6 + 3√3
<u>Explanation:</u>
Area of the square = 3
a X a = 3
a² = 3
a = √3
Therefore, QR, RS, SP, PQ = √3
ΔBAC ≅ ΔBQR
Therefore,
In ΔBAC, BA = AC = BC because the triangle is equilateral
So,
BQ = √3
So, BQ, QR, BR = √3 (equilateral triangle)
Let AP and SC be a
So, AQ and RC will be 2a
In ΔAPQ,
(AP)² + (QP)² = (AQ)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
Similarly, in ΔRSC
(SC)² + (RS)² = (RC)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
So, AP and SC = 1
and AQ and RC = 2 X 1 = 2
Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR
Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3
Perimeter = 6 + 3√3
Therefore, the perimeter of the triangle is 6 + 3√3
Answer:
a) H0 : P = 0.07
Ha : P ≠ 0.07
b) p -value = 0.1770
Step-by-step explanation:
P = 7% = 0.07
x (result ) = 7 , n = 163
p = x / n = 7 / 163 = 0.043
<u>a) Using a significance level ( ∝ ) of 0.05, estimate the appropriate hypothesis</u>
H0 : P = 0.07
Ha : P ≠ 0.07
conduct a Z- test statistic
Z = ( p - P ) /
= ( 0.043 - 0.07 ) / = - 1.35
Critical value ( Z₀.₀₂₅ ) = ± 1.96
<em>we fail to reject H0 given that | z | < Zcritical because there is not enough evidence to conclude that proportion change</em>
<u>b) Determine the p-value of the test </u>
P-value = P ( | Z | > Z )
= 2 * P ( Z < -1.35 )
= 0.1770
The p-value ( 0.1770 ) > ∝ ( 0.05 ) hence we fail to reject H0 ( i.e. the conclusion agrees with part a above )