The wind-chill index W is the perceived temperature when the actual temperature is T and the wind speed is v, so we can write W
= f(T, v). (a) Estimate the values of fT(−15, 40) and fv(−15, 40). (Round your answers to two decimal places.) fT(−15, 40) ≈ fv(−15, 40) ≈ What are the practical interpretations of these values? When the actual temperature is −15°C and the wind speed is 40 km/h, the apparent temperature ---Select--- by about °C for every degree that the actual temperature rises. When the actual temperature is −15°C and the wind speed is 40 km/h, the apparent temperatur
1 answer:
Answer:
hello your question is incomplete attached below is the complete question
estimating: Ft( -15,40 ) = ( 1.2 + 1.4 ) / 2 = 1.30⁰c
when the actual temperature is -15⁰c and the wind speed is 40 km/h the apparent temperatures increase by 1.3⁰c that the actual temperature rises
estimating: Fv ( -15,40 ) = (-0.2 + -0.1 ) / 2 = -0.15⁰c
when the actual temperature -15⁰c and the wind speed is 40 km/h the apparent temperature decreases by about 0.15⁰c for every km/h that the wind speed
Step-by-step explanation:
A) Estimating the values of Ft(−15, 40) and fv(−15, 40)
attached below is a detailed solution
To estimate the value of Ft( -15,40 ) we have to take an average value hence Ft ( -15,40 ) = ( 1.2 + 1.4 ) / 2 = 1.30
and this means that when the actual temperature is -15⁰c and the wind speed is 40 km/h the apparent temperatures increase by 1.3⁰c that the actual temperature rises
To estimate the value of Fv(-15,40 ) we have to take an average value
hence Fv ( -15,40 ) = (-0.2 + -0.1 ) / 2 = -0.15
and this means that when the actual temperature -15⁰c and the wind speed is 40 km/h the apparent temperature decreases by about 0.15⁰c for every km/h that the wind speed
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Probability of a student accepted at both Dartmouth and Harvard is:
50% x 40% = 20%
Probability of a student accepted at Dartmouth but not Harvard is:
50% x 60% = 30%
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