A briefcase has a three-digit lock code that does not include zero as a digit.

Mathematics

2 answers:

Strike441 [17]3 years ago

6 0

The correct answer is:

A. 24 out of 504.

The probability that the lock code consists of all even digits is 24 out of 504, if the same digit is not used more than once in the lock code.

$|Huntrw6|$

Sladkaya [172]3 years ago

4 0

First it can be a 2,4,6,8. Four numbers.
Then one is removed, leaving three numbers.
Then one is removed, leaving two numbers.
Possibilities for that to happen: 4*3*2 = 24.
Total amount of possibilities: 9*8*7 = 504.
So the answer is A: 24 out of 504.

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Answer:

The correct options are;

1. Definition of supplementary angles

2. m∠1 + m∠2 = m∠1 + m∠3

3. m∠2 = m∠3

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Step-by-step explanation:

The two column proof is presented as follows;

Statement ${}$ Reason

1. ∠1 and ∠2 are supplementary ${}$ Given

${}$ ∠1 and ∠3 are supplementary

2. m∠1 + m∠2 = 180° ${}$ Definition of supplementary angles

${}$ m∠1 + m∠3 = 180°

3. m∠1 + m∠2 = m∠1 + m∠3 ${}$ Transitive Property

4. m∠2 = m∠3 ${}$ Subtraction Property of Equality

5. ∠2 ≅ ∠3 ${}$ Definition of Congruent Angles

Given that angles ∠1 and ∠2 are supplementary angles and angles ∠1 and ∠3 are are also supplementary angles, then the sums of m∠1 + m∠2 and m∠1 + m∠3 are equal, therefore, ∠2 and ∠3 have equal quantitative value and therefore ∠2 = ∠3 and by definition, ∠2 ≅ ∠3.

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