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Calculate the final temperature and the amount of maize distributed to customers.

Mathematics

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3. Determine the final temperature of the patient.

  1. Start with the initial temperature.

    Initial temperature=39∘C\text{Initial temperature} = 39^\circ\text{C}
  2. Subtract the temperature fall.

    39∘C−3∘C=36∘C39^\circ\text{C} - 3^\circ\text{C} = 36^\circ\text{C}
  3. Add the temperature increase.

    36∘C+1∘C=37∘C36^\circ\text{C} + 1^\circ\text{C} = 37^\circ\text{C}

    The final temperature of the patient is 37∘C\boxed{37^\circ\text{C}}.

    4. How many kilograms of maize did each customer receive?

  4. Calculate the initial total mass of maize.

    Initial bags=350\text{Initial bags} = 350 Weight per bag=90 kg\text{Weight per bag} = 90 \text{ kg} Initial total mass=350 bags×90 kg/bag=31500 kg\text{Initial total mass} = 350 \text{ bags} \times 90 \text{ kg/bag} = 31500 \text{ kg}
  5. Calculate the total bags sold.

    Bags sold day 1=120\text{Bags sold day 1} = 120 Bags sold day 2=50\text{Bags sold day 2} = 50 Total bags sold=120+50=170 bags\text{Total bags sold} = 120 + 50 = 170 \text{ bags}
  6. Calculate the bags remaining after sales.

    Bags remaining=350−170=180 bags\text{Bags remaining} = 350 - 170 = 180 \text{ bags}
  7. Calculate the total bags after purchasing more.

    Bags purchased=80\text{Bags purchased} = 80 New total bags=180+80=260 bags\text{New total bags} = 180 + 80 = 260 \text{ bags}
  8. Calculate the total mass of maize available for customers.

    Total mass for customers=260 bags×90 kg/bag=23400 kg\text{Total mass for customers} = 260 \text{ bags} \times 90 \text{ kg/bag} = 23400 \text{ kg}
  9. Divide the total mass by the number of customers.

    Number of customers=5\text{Number of customers} = 5 Mass per customer=23400 kg5=4680 kg\text{Mass per customer} = \frac{23400 \text{ kg}}{5} = 4680 \text{ kg}

    Each customer received 4680 kg\boxed{4680 \text{ kg}} of maize.

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