Based on the graph, find the range of y = f(x).
-The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is $$W ( t ) = \left\{ \begin{array} { l l }
t & \text { if } 0 \leq v

Question

Based on the graph, find the range of y = f(x).
-The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is $$W ( t ) = \left\{ \begin{array} { l l } 
t & \text { if } 0 \leq v < 1.79 \
33 - \frac { ( 10.45 + 10 \sqrt { v } - v ) ( 33 - t ) } { 22.04 } & \text { if } 1.79 \leq \mathrm { v } < 20 \
33 - 1.5958 ( 33 - t ) & \text { if } v \geq 20
\end{array} \right.$$ where v represents the wind speed (in meters per second) and t represents the air temperature (°C). Compute the wind chill for an air temperature of 15°C and a wind speed of 12 meters per second. (Round the answer to one decimal place.)

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The Answer of Based on the graph, find the range of y = f(x).
-The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is $$W ( t ) = \left\{ \begin{array} { l l } 
t & \text { if } 0 \leq v < 1.79 \
33 - \frac { ( 10.45 + 10 \sqrt { v } - v ) ( 33 - t ) } { 22.04 } & \text { if } 1.79 \leq \mathrm { v } < 20 \
33 - 1.5958 ( 33 - t ) & \text { if } v \geq 20
\end{array} \right.$$ where v represents the wind speed (in meters per second) and t represents the air temperature (°C). Compute the wind chill for an air temperature of 15°C and a wind speed of 12 meters per second. (Round the answer to one decimal place.)

Based on the Graph, Find the Range of Y =