Answer:
If I'm correct, the answer should be negative 40/3.
Step-by-step explanation:
Because you first combine multiplied terms into a single fraction, then you multiply all terms by the same value to eliminate fraction denominators. Now you simplify it, by canceling multiplied terms that are in the denominator and then you multiply the numbers. Now you will divide both sides of the equation by the same term. Finally, you will simplify the equation.
(Hope this helps) I had to get a little help myself on this problem)
Answer:
541.67m²
Step-by-step explanation:
Step 1
We find the third angle
Sum of angles in a triangle = 180°
Third angle = Angle V = 180° - (63 + 50)°
= 180° - 113°
Angle V = 67°
Step 2
Find the sides x and v
We find these sides using the sine rule
Sine rule or Rule of Sines =
a/ sin A = b/ Sin B
Hence for triangle VWX
v/ sin V = w/ sin W = x / sin X
We have the following values
Angle X = 50°
Angle W = 63°
Angle V = 67°
We are given side w = 37m
Finding side v
v/ sin V = w/ sin W
v/ sin 67 = 37/sin 63
Cross Multiply
sin 67 × 37 = v × sin 63
v = sin 67 × 37/sin 63
v = 38.22495m
Finding side x
x / sin X= w/ sin W
x/ sin 50 = 37/sin 63
Cross Multiply
sin 50 × 37 = v × sin 63
x = sin 50 × 37/sin 63
x = 31.81082m
To find the area of triangle VWX
We use heron formula
= √s(s - v) (s - w) (s - x)
Where S = v + w + x/ 2
s = (38.22 + 37 + 31.81)/2
s = 53.515
Area of the triangle = √53.515× (53.515 - 38.22) × (53.515 - 37 ) × (53.515 - 31.81)
Area of the triangle = √293402.209
Area of the triangle = 541.66614164081m²
Approximately to the nearest tenth = 541.67m²
Answer:
Answer: 155.12km
Step-by-step explanation:
sin(64˚) x 78 = 70.11 (2DP)
70.12 (2DP) + 85 = 155.1059356 (155.12 - 2DP)
Answer:
a) 151lb.
b) 6.25 lb
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean 
and standard deviation 
, a large sample size can be approximated to a normal distribution with mean and standard deviation 
.
In this problem, we have that:
So
a) The expected value of the sample mean of the weights is 151 lb.
(b) What is the standard deviation of the sampling distribution of the sample mean weight?
This is 
