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Indicate if the following statement is true or false:A * 3 >= B – 14 \ 3 AND F OR C <= D + 3 where A = 7, B = 9, C = 14, D = -4 and F = FALSETrueFalse

Question

Indicate if the following statement is true or false:

A * 3 >= B – 14

3 AND F OR C <= D + 3

where

A = 7,
B = 9,
C = 14,
D = -4

and F = FALSE

True
False

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Solution

To determine whether the statement is true or false, we will evaluate it step by step using the provided values for A A , B B , C C , D D , and F F .

1. Break Down the Problem

The original expression is: A⋅3≥B−143 AND F OR C≤D+3 A \cdot 3 \geq B - \frac{14}{3} \text{ AND } F \text{ OR } C \leq D + 3

We need to evaluate the following parts:

  1. Calculate A⋅3 A \cdot 3
  2. Calculate B−143 B - \frac{14}{3}
  3. Check the condition A⋅3≥B−143 A \cdot 3 \geq B - \frac{14}{3}
  4. Evaluate F F
  5. Calculate D+3 D + 3
  6. Check the condition C≤D+3 C \leq D + 3
  7. Combine results using logical operations.

2. Relevant Concepts

  • Basic arithmetic operations
  • Logical operations (AND, OR)

3. Analysis and Detail

  1. Calculate A⋅3 A \cdot 3 A⋅3=7⋅3=21 A \cdot 3 = 7 \cdot 3 = 21

  2. Calculate B−143 B - \frac{14}{3} 143=4.67  ⟹  B−143=9−4.67≈4.33 \frac{14}{3} = 4.67 \implies B - \frac{14}{3} = 9 - 4.67 \approx 4.33

  3. Evaluate the condition A⋅3≥B−143 A \cdot 3 \geq B - \frac{14}{3} 21≥4.33(This is TRUE) 21 \geq 4.33 \quad \text{(This is TRUE)}

  4. Evaluate F F F=FALSE F = \text{FALSE}

  5. Calculate D+3 D + 3 D+3=−4+3=−1 D + 3 = -4 + 3 = -1

  6. Check the condition C≤D+3 C \leq D + 3 C=14  ⟹  14≤−1(This is FALSE) C = 14 \implies 14 \leq -1 \quad \text{(This is FALSE)}

  7. Combine results using logical operations Condition: (TRUE AND FALSE) OR FALSE  ⟹  TRUE AND FALSE=FALSEthen FALSE OR FALSE=FALSE \text{Condition: (TRUE AND FALSE) OR FALSE} \implies \text{TRUE AND FALSE} = \text{FALSE} \quad \text{then } \text{FALSE OR FALSE} = \text{FALSE}

4. Verify and Summarize

The values evaluated indicate that the final result of the logical expression is FALSE.

Final Answer

FALSE

This problem has been solved

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