Applying the knowledge of fractions, the bigger one is: one-sixth of 138.
<h3>What are Fractions?</h3>
Fractions can be defined as a part of a digit or number, and are not whole numbers. For example, a third is 1/3, one-sixth is 1/6, which are both fractions.
Therefore, let's evaluate each statement:
1/3 of 45 = 1/3 × 45 = 15
1/6 of 138 = 1/6 × 138 = 23
Therefore, the bigger of the two is: one-sixth of 138.
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Answer:
-4
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
4x + 5y = -12
-2x + 3y = -16
<u>Step 2: Rewrite Systems</u>
-2x + 3y = -16
- Multiply everything by 2: -4x + 6y = -32
<u>Step 3: Redefine Systems</u>
4x + 5y = -12
-4x + 6y = -32
<u>Step 4: Solve for </u><em><u>y</u></em>
<em>Elimination</em>
- Combine 2 equations: 11y = -44
- Divide 17 on both sides: y = -4
Answer:
55.04 m 2
Step-by-step explanation:
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By definition of complement,
Pr[not P | G and T] = 1 - Pr[P | G and T]
and by definition of conditional probability,
Pr[not P | G and T] = 1 - Pr[P and G and T] / Pr[G and T]
Pr[not P | G and T] = 1 - (16/100) / (33/100)
Pr[not P | G and T] = 1 - 16/33
Pr[not P | G and T] = 17/33
Answer:
a, (x^2)^3 = x^6 = x^2.x^4
b, x^5·x^7 = x^12 = (x^3)^4
c, x^4·x^22 = x^26 = (x^2)^13
d, (x^2)^8 = x^16 = x^2.x^14
e, x^10/x^3 = x^7 = x^20.x^(-13)
f, x^(-3) = x^4.x^(-7)
g, 1/x^(-3) = x^3 = x^4.x^(-1)
Hope this helps!
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