Answer:
1 plane
Step-by-step explanation:
Let's suppose the number of planes waiting or on the runway " P "
the number of planes taking off per hour ' 
'
the time for waiting and the runway 
so:
P = x
= 18 airplanes per hour
we know that 1 min = 60s
36s = 36/60 = 0.6 min
Also, 3 min and 30 s = 3 + 30/60 = 3.5 min
Next to find the time for waiting and the runway
∴= 0.6 + 3.5 = 4.1 min/60 (converting into hour)
= 0.068 hour
P = 18x0.068 = 1.23
therefore, there is 1 plane either on the runway or waiting to take off
So, there is 1 plane either on the runway or waiting to take off
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Answer:
A=12
B=54
C=18
Step-by-step explanation:
A=first
B=second
C=third
A+B+C=84
B=3C
A=C-6
C-6+3C+C=84
5C-6=84
5C=90
C=18
A=C-6
=(18)-6
=12
B=3C
=3(18)
=54
Answer:
1. E(Y) = 50.54°F
2. SD(Y) = 11.34°F
Step-by-step explanation:
We are given that The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3°C with a standard deviation SD(X) = 3.5°C.
The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.
(1) Y = (9/5)X + 32
E(Y) = E((9/5)X + 32) = E((9/5)X) + E(32)
= (9/5) * E(X) + 32 {
expectation of constant is constant}
= (9/5) * 10.3 + 32 = 50.54
Therefore, E(Y), the expected daily high in Montreal during April in degrees Fahrenheit is 50.54°F .
(2) Y = (9/5)X + 32
SD(Y) = SD((9/5)X + 32) = SD((9/5)X) + SD(32)
= 
* SD(X) + 0 { standard deviation of constant is zero}
= 
* 3.5 = 11.34°F
Therefore, SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit is 11.34°F .
52,458<52,485
52,485-52,458=27
Answer:
Angle 1: 95 degrees
Angle 2: 85 degrees
Angle 3: 95 degrees
Step-by-step explanation:
Vertical angles are opposite and equal so in this case angle 1 and 3 are vertical and so are 2 and 85°
So that means angle 2 also equals 85 degrees and because of supplementary angles the other angle must add up to equal 180 degrees
85 + ? = 180
?=95
So both angle 1 and 3 equal 95 degrees
