You are planning to take to a trip to Montreal, Canada during the month of April and you want to bring clothing that is appropri
1 answer:
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 .
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Answer:
Option D.
Step-by-step explanation:
The area of the circular bases is:
Where
is the radius of the circle
Then
The area of the rectangle is:
Where
b is the width of the rectangle and is the length
Then the area of the rectangle is:
Finally the total area is:
Answers:
x = angle MLN = 35 degrees
angle FLJ = 55 degrees
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Explanation:
This problem is fairly tricky if you're not sure what to look for. A slight clue is that they've marked a red point that strangely doesn't have a label on it. It's a fairly small point but it's definitely there if you look closely. While this point's location isn't exactly what we want, we're fairly close. Start at point L and draw a ray through the center point P. Ray LP will intersect the circle at point A, as shown in the diagram below.
From here, draw segments FA and MA. Now notice that inscribed angle LMF = 55 and inscribed angle FAL both subtend the same arc. The term "subtend" basically means "cut off". This arc that the inscribed angles subtend is minor arc FL. A minor arc is where you travel the shorter path around the circle, which indicates its measure is less than 180 degrees.
Since inscribed angles LMF and FAL subtend the same minor arc, this makes the inscribed angles to be congruent.
In short: angle FAL is 55 degrees
----------------------
Segment LA goes through the center P. Through Thales theorem, we know that inscribed angle LFA is 90 degrees. Consequently, we can determine that inscribed angle FLA is 90-55 = 35 degrees.
Segment JL is tangent to the circle, meaning that angle ALJ is 90 degrees. So angle FLJ is 90-35 = 55 degrees.
It's not a coincidence that angle FLJ, angle FAL, and angle LMF are the same measure.
----------------------
We found that angle FAL was 55 degrees. Applying Thales theorem again shows that angle MAF is 90 degrees. Therefore, angle LAM is 90-55 = 35 degrees.
Focus now on triangle LMA. This is also a right triangle (Thales Theorem). The upper acute angle we found was angle LAM = 35, so the lower acute angle is ALM = 55.
Then we can find angle MLN = (angle ALN) - (angle ALM) = 90 - 55 = 35
In short, angle MLN = 35 degrees
Similar to the previous section, it is not a coincidence that angles LFM, LAM and MLN are the same measure.
------------------------
As an alternative, since we know angle FLJ = 55, and angle FLM is 90 degrees, this means...
(angle FLJ)+(angle FLM) + (angle MLN) = 180
55 + 90 + angle MLN = 180
145 + angle MLN = 180
angle MLN = 180 - 145
angle MLN = 35 degrees
