[Solved] Given the code below, what will be printed if sensor_value = 75? def check_sensor(value): if value > 70: print("Alert:... Given the code below, what will be printed if sensor_value = 75? | Raspberry Pi

Raspberry Pi - Automation and Scheduling

Given the code below, what will be printed if sensor_value = 75?

def check_sensor(value):
    if value > 70:
        print("Alert: High sensor value!")
    else:
        print("Sensor value normal.")

check_sensor(sensor_value)

AAlert: High sensor value!

BNo output

CSensor value normal.

DSyntaxError

Step-by-Step Solution

Solution:

Step 1: Analyze the condition with sensor_value = 75
Since 75 > 70, the if condition is true.
Step 2: Determine which print statement runs
The code prints "Alert: High sensor value!" because the if condition is met.
Final Answer:
Alert: High sensor value! -> Option A
Quick Check:
75 > 70 triggers alert print [OK]

Quick Trick: Check if condition true or false to find printed message [OK]

Common Mistakes:

MISTAKES

Confusing else output with if output
Assuming no output if condition true
Thinking code has syntax errors

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Given the code below, what will be printed if sensor_value = 75?

Step 1: Analyze the condition with sensor_value = 75

Step 2: Determine which print statement runs

Final Answer:

Quick Check:

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