Answer:
UV = 4.5 (not rounded)
TV = 3√3
m∠U = 60˚
Step-by-step explanation:
UV = Use sin function since you’re finding the hypotenuse with a given angle and opposite side.
sin(30) = 9/UV
9sin(30) = UV
4.5 = UV
TV = Use Tangent function (opp. over adj.) since you’re finding the adjacent side with a given angle and opposite side.
tan(30) = 9/TV
9tan(30) = TV
3√3 = TV
m∠U = Sum of all interior angles of a triangle is 180˚:
90˚ + 30˚ = 120˚
180˚ - 120˚ = 60˚
Hope this helps :)
Answer:
The equation of the required line is y = x + 5
Step-by-step explanation:
The equation of the given line is y = x - 2
The required line = A line parallel to the given line
The point through which the required line passes = (-3, 2)
The general form of the equation of a straight line, is y = m·x + c
Where;
m = The slope of the line
By comparison, the slope of the given line, m = 1
When two lines are parallel, their slope are equal
Therefore, the slope of the required line = m = 1
The equation of the required line in point and slope form is therefore;
y - 2 = x - (-3) = x + 3
∴y = x + 3 + 2 = x + 5
The equation of the required line is therefore;
y = x + 5.
Answer:
Probability that a customer pays the bill by mail, given that the customer is female is 0.20.
Step-by-step explanation:
The complete question is:
Online By Mail Total
Male 12 8 20
Female 24 6 30
Total 36 14 50
A utility company asked 50 of its customers whether they pay their bills online or by mail. What is the probability that a customer pays the bill by mail, given that the customer is female.
Probability that a customer pays the bill by mail, given that the customer is female is given by = P(Pays bill by mail / Customer is female)
P(Pays bill by mail/Customer is female) = 
Now, Probability that customer is female = 
Also, Probability that customer pays bill by mail and is female =
So, Required probability = 
= 
= 
= 0.20
Hence, probability that a customer pays the bill by mail, given that the customer is female is 0.20.
If he fills 5.9 bags for 18 hours, he will have filled a total of 106.2 bags
