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23 Jun 2025, 08:24
Kudos
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
50% (00:31) correct
50% (00:23) wrong based on 6 sessions
History
Date
Time
Result
Not Attempted Yet
At 3:00 A.M. the temperature was 13° below zero. By noon it had risen to 32°. What was the average hourly increase in temperature?
A. \((\frac{19}{9})°\)
B. \((\frac{19}{6})°\)
C. \(5°\)
D. \(7.5°\)
E. \(45°\)
A. \((\frac{19}{9})°\)
B. \((\frac{19}{6})°\)
C. \(5°\)
D. \(7.5°\)
E. \(45°\)
24 Jun 2025, 22:12
Kudos
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OE
In the 9 hours from 3:00 to 12:00, the temperature rose 32 – (–13) = 32 + 13 = 45 degrees. So, the average hourly increase was 45° ÷ 9 = 5°.
Answer: C
In the 9 hours from 3:00 to 12:00, the temperature rose 32 – (–13) = 32 + 13 = 45 degrees. So, the average hourly increase was 45° ÷ 9 = 5°.
Answer: C
