Answer:
3/8
Step-by-step explanation:
well there are common denominators so all you have to do is subtract 5 and 2
S/<span>πr +r =l
might be -r
</span>
Answer:
Step-by-step explanation:
42:7=ED:24
cross multiply 42 into 24.
Then divide by 7
you will get ED
Then use pythageorous theorem.
Ed square + 42 square=Ef square
Then do square root of Ef square
Answer:
(-4, -8)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
x - 2y = 12
5x + 3y = -44
<u>Step 2: Rewrite Systems</u>
x - 2y = 12
- [Multiplication Property of Equality] Multiply everything by -5: -5x + 10y = -60
<u>Step 3: Redefine Systems</u>
-5x + 10y = -60
5x + 3y = -44
<u>Step 4: Solve for </u><em><u>y</u></em>
<em>Elimination</em>
- Combine 2 equations: 13y = -104
- [Division Property of Equality] Divide 13 on both sides: y = -8
<u>Step 5: Solve for </u><em><u>x</u></em>
- Define original equation: x - 2y = 12
- Substitute in <em>y</em>: x - 2(-8) = 12
- Multiply: x + 16 = 12
- [Subtraction Property of Equality] Subtract 16 on both sides: x = -4
Answer:
0.5122 or 51.22%
Step-by-step explanation:
In a certain city, in June Probability of cloudy days = P(cloudy) = 0.41
Probability of cloudy and rainy = P(cloudy and rainy) = 0.21
Probability of rainy if we already know it is cloudy = ![\frac{\text{[P(cloud and rainy)]}}{[P(cloud)]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%5BP%28cloud%20and%20rainy%29%5D%7D%7D%7B%5BP%28cloud%29%5D%7D)
= 
= 0.512195122 ≈ 0.5122
Therefore, the probability that a randomly selected day in June will be rainy if it is cloudy is 0.5122 or 51.22%
