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On a 
40^(@)F day, a 25 mile per hour 
(mph) wind creates a wind chill of 
29^(@)F. To approximate the wind chill at various temperatures caused by a 
25mph wind, Elon uses the following rule: take 24 less than 
(4)/(3) of the air temperature 
(^(@)F). If a 
25mph wind creates a wind chill of 
-5^(@)F, which of the following best approximates the corresponding air temperature?
Choose 1 answer:
(A) 
-30.7^(@)F
(B) 
14.3^(@)F
(C) 
21.7^(@)F
(D) 
25.3^(@)F

On a 40F 40^{\circ} \mathrm{F} day, a 2525 mile per hour (mph) (\mathrm{mph}) wind creates a wind chill of 29F 29^{\circ} \mathrm{F} . To approximate the wind chill at various temperatures caused by a 25mph 25 \mathrm{mph} wind, Elon uses the following rule: take 2424 less than 43 \frac{4}{3} of the air temperature (F) \left({ }^{\circ} \mathrm{F}\right) . If a 25mph 25 \mathrm{mph} wind creates a wind chill of 5F -5^{\circ} \mathrm{F} , which of the following best approximates the corresponding air temperature?\newlineChoose 11 answer:\newline(A) 30.7F -30.7^{\circ} \mathrm{F} \newline(B) 14.3F 14.3^{\circ} \mathrm{F} \newline(C) (mph) (\mathrm{mph}) 00\newline(D) (mph) (\mathrm{mph}) 11

Full solution

Q. On a 40F 40^{\circ} \mathrm{F} day, a 2525 mile per hour (mph) (\mathrm{mph}) wind creates a wind chill of 29F 29^{\circ} \mathrm{F} . To approximate the wind chill at various temperatures caused by a 25mph 25 \mathrm{mph} wind, Elon uses the following rule: take 2424 less than 43 \frac{4}{3} of the air temperature (F) \left({ }^{\circ} \mathrm{F}\right) . If a 25mph 25 \mathrm{mph} wind creates a wind chill of 5F -5^{\circ} \mathrm{F} , which of the following best approximates the corresponding air temperature?\newlineChoose 11 answer:\newline(A) 30.7F -30.7^{\circ} \mathrm{F} \newline(B) 14.3F 14.3^{\circ} \mathrm{F} \newline(C) (mph) (\mathrm{mph}) 00\newline(D) (mph) (\mathrm{mph}) 11
  1. Understand Wind Chill Rule: Understand the wind chill approximation rule given by Elon.\newlineThe rule is to take 2424 less than (43)(\frac{4}{3}) of the air temperature (°F).\newlineMathematically, this can be represented as:\newlineWind chill = (43)×(\frac{4}{3}) \times Air temperature 24- 24
  2. Set Up Equation: Set up the equation using the wind chill value provided in the problem.\newlineWe are given that the wind chill is 5-5°F, so we plug this value into the equation:\newline5=43×Air temperature24-5 = \frac{4}{3} \times \text{Air temperature} - 24
  3. Solve for Temperature: Solve for the air temperature.\newlineFirst, add 2424 to both sides of the equation to isolate the term with the air temperature:\newline5+24=(43)×Air temperature-5 + 24 = \left(\frac{4}{3}\right) \times \text{Air temperature}\newline19=(43)×Air temperature19 = \left(\frac{4}{3}\right) \times \text{Air temperature}
  4. Multiply by Reciprocal: Multiply both sides of the equation by the reciprocal of (43)(\frac{4}{3}) to solve for the air temperature.\newlineAir temperature = 19×(34)19 \times (\frac{3}{4})\newlineAir temperature = 574\frac{57}{4}\newlineAir temperature = 14.2514.25
  5. Match to Answer Choice: Match the calculated air temperature to the closest answer choice.\newlineThe calculated air temperature is 14.25°F14.25\degree F, which is closest to answer choice (B) 14.3°F14.3\degree F.

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