QUESTION IMAGE
Question
- using the following variable names and assignment condition complete the truth table that captures the functionality of this audio alarm. once complete take a picture and upload below.
d: door → 0 = door open / 1 = door close
k: key → 0 = key not in ignition / 1 = key in ignition
s: seat belt → 0 = not buckled / 1 = buckled
b: buzzer → 0 = buzzer off / 1 = buzzer on
logic statement: the buzzer is on whenever the door is open or when the key is in the ignition and the seat belt is not buckled.
| d (door) | k (key) | s (seat belt) | b (buzzer) |
|---|---|---|---|
| 0 | 0 | 1 | |
| 0 | 1 | 0 | |
| 0 | 1 | 1 | |
| 1 | 0 | 0 | |
| 1 | 0 | 1 | |
| 1 | 1 | 0 | |
| 1 | 1 | 1 |
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Step1: Translate logic to Boolean formula
The buzzer \( B \) is on when \( D=0 \) (door open) OR (\( K=1 \) AND \( S=0 \)). In Boolean terms:
\( B = \overline{D} \lor (K \land \overline{S}) \)
Step2: Calculate B for each row
Row1: \( D=0, K=0, S=0 \)
\( B = 1 \lor (0 \land 1) = 1 \lor 0 = 1 \)
Row2: \( D=0, K=0, S=1 \)
\( B = 1 \lor (0 \land 0) = 1 \lor 0 = 1 \)
Row3: \( D=0, K=1, S=0 \)
\( B = 1 \lor (1 \land 1) = 1 \lor 1 = 1 \)
Row4: \( D=0, K=1, S=1 \)
\( B = 1 \lor (1 \land 0) = 1 \lor 0 = 1 \)
Row5: \( D=1, K=0, S=0 \)
\( B = 0 \lor (0 \land 1) = 0 \lor 0 = 0 \)
Row6: \( D=1, K=0, S=1 \)
\( B = 0 \lor (0 \land 0) = 0 \lor 0 = 0 \)
Row7: \( D=1, K=1, S=0 \)
\( B = 0 \lor (1 \land 1) = 0 \lor 1 = 1 \)
Row8: \( D=1, K=1, S=1 \)
\( B = 0 \lor (1 \land 0) = 0 \lor 0 = 0 \)
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Completed truth table (B column values, top to bottom):
1, 1, 1, 1, 0, 0, 1, 0
Full formatted truth table:
| D (Door) | K (Key) | S (Seat Belt) | B (Buzzer) |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |