A=43°
B=82°
c=28
1) A+B+C=180°
Replacing A=43° and B=82° in the equation above:
43°+82°+C=180°
125°+C=180°
Solving for C. Subtracting 125° both sides of the equation:
125°+C-125°=180°-125°
C=55° (option B or C)
2) Law of sines
a/sin A=b/sin B=c/sin C
Replacing A=43°, B=82°, C=55°, and c=28 in the equation above:
a/sin 43°=b/sin 82°=28/sin 55°
2.1) a/sin 43°=28/sin 55°
Solving for a. Multiplying both sides of the equation by sin 43°:
sin 43°(a/sin 43°)=sin 43°(28/sin 55°)
a=28 sin 43° / sin 55°
Using the calculator: sin 43°=0.681998360, sin 55°=0.819152044
a=28(0.681998360)/0.819152044
a=23.31185549
Rounded to one decimal place
a=23.3
2.2) b/sin 82°=28/sin 55°
Solving for a. Multiplying both sides of the equation by sin 82°:
sin 82°(b/sin 82°)=sin 82°(28/sin 55°)
b=28 sin 82° / sin 55°
Using the calculator: sin 82°=0.990268069, sin 55°=0.819152044
b=28(0.990268069)/0.819152044
b=33.84903466
Rounded to one decimal place
b=33.8
Answer: Option B) C=55°, b=33.8, a=23.3
The graph is the one with coordinates (1,2) as the center.
300,000 <-- three hundred thousand
5,000 <-- five thousand
63 <-- sixty three
300,000
5,000
63
305,063
305,063 is your answer.
Hope this helps.
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560



P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = 
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)


From z tables
P(Y > 20)
0.0903