You are planning to take to a trip to Montreal, Canada during the month of April and you want to bring clothing that is appropriate for the weather. The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3oC with a standard deviation SD(X) = 3.5oC. You want to convert these Celsius temperatures to oF (degrees Fahrenheit). The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.
1. What is E(Y), the expected daily high in Montreal during April in degrees Fahrenheit?
2. What is SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit?

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 .

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

the car was traveling 40 mph before it started skidding. ( 2 points ) 17. was the driver were her brakes working at 75% efficiency or better? ( 2 points ) yes 29.

step-by-step explanation: