As per the given question,
Temperature reading on faharenheit scale =28
We know that C=5(F−32)9C=\dfrac{5(F-32)}{9}C=95(F−32)
C=5(28−32)9=209∘CC=\dfrac{5(28-32)}{9}=\dfrac{20}{9}^\circ CC=95(28−32)=920∘C
Flow diagram is given below,
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