The answer is 11/36
2/12 chance of rolling fours
because there are 2 sides containing a four on both dice combined and 12 sides in total.
Doubles mean you have to roll the same number simultaneously so let’s say we want to calculate the probability for double ones: then it’s 1/6 on the first dice for a one, and 1/6 on the second dice to land on a one as well.
I personally like to imagine a box like this:
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If you have one dice then it’s just a random segment on one of the lines. If you want the specific result from two dice then you want two specific segments which is also the 1 specific tile out of 36 (6 width times 6 height). So you multiply.
1/6 * 1/6 = 1/36 chance to roll double of ones
And 1/36 chance to roll double twos, threes, fours, fives, and sixes. But we don’t count the double fours because any four will do. So:
1/36 * 5 = 5/36
So for the probability of either doubles or containing a four is the probability of doubles of either number plus the probability of either dice being a four:
5/36 + 2/12 =
5/36 + 6/36 =
11/36
1.
v x w = 8 i - 24 j + 15 k + 10 j - 12 i - 24 k =
= - 4 i - 14 j - 9 k
Answer: D )
2.
v x w = -16 i - 16 j - 18 k + 12 j + 48 i - 8 k =
= 32 i - 4 j - 10 k
Answer: C )
3.
The cross product:
< - 6, 7, 2 > x < 8, 5, -3 > =
= - 21 i + 16 j - 30 k - 18 j - 10 i - 56 k =
= - 31 j - 2 j - 86 k = < - 31, - 2, - 86 >
Vectors are perpendicular if: cos ( u, v ) = 0
< - 6 , 7. 2 > * < -31, - 2, - 86 > = 186 - 14 - 172 = 0
< 8, 5 , - 3 > * < - 31, - 2, - 86 > = -248 - 10 + 258 = 0
Answer: A ) < - 31, - 2 , - 86 >, yes.
Answer:
the answer is -6,5
Step-by-step explanation:
just look at the left side ( the x axis )first then right(the y axis)
Answer:
59.15 in^2
Step-by-step explanation:
Using Heron's formula
semiperimeter , s = 1/2 ( 28.8+18+12) = 29.4
Area = sqrt ( 29.4(29.4-28.8)(29.4-18)(29.4-12) ) = 59.15 in^2
