The distributive property is a(b + c) = ab + ac
Answer:
The frequency of rolling a 3 or a 6 would be the same even if you try thousands or millions times.
The probability of rolling a 3 is: 1/ total outcomes probability = 1/6
The probability of rolling a 6 is: 1/ total outcomes probability = 1/6
The probability of rolling a 6 or a 3 = 2/ total outcomes probability = 2/6 = 1/3
Therefore, if you rolling the cube 600 times, the probability of rolling a 3 or a 6 is still 1/6. If 3 and 6 are both allowed, the probability would become 1/3
Hope this helped :3
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
Answer:
Step 3
Step-by-step explanation:
7( 8.5-1.5) +8 ÷2
PEMDAS
Parentheses
7*7 +8 ÷2
Then multiplication and division from left to right
49 +8 ÷2
We are still on multiplication and division
49+4
So step 3 is the first mistake
