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
Answer:
area of rhombus =1/2 d1×d2
20=1/2 ×10×d2
d2=4
intersection of diagonal make right angle and cut in equal parts
d1/2=10/2=5
d2/2=4/2=2
use Pythagoras in right angle to bide side
side^2= 5^2 +2^2
side = √29
Answer:
100% increase
Step-by-step explanation:
Calculate percentage change
from V1 = 20 to V2 = 40
(V2−V1)|V1|×100
=(40−20)|20|×100
=2020×100
=1×100
=100%change
=100%increase
Answer:
Step-by-step explanation:
given is a system of linear equations in 3 variables as
This can be represented in matrix form as
AX=B Or
![\left[\begin{array}{ccc}-1&-4&2\\1&2&-1\\1&1&-1\end{array}\right] *\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-10\\11\\14\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-4%262%5C%5C1%262%26-1%5C%5C1%261%26-1%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-10%5C%5C11%5C%5C14%5Cend%7Barray%7D%5Cright%5D)
So solution set
X would be 
|A|=-1(-1)+4(0)+2(-1)=--1
Cofactors of A are
-1 0 -1
-2 -1 -3
0 1 2
So inverse of A is
1 2 0
0 1 -1
1 3 -2
Solution set would be
x=12
y=-3
z=-5
Okay I can help you with your math!!
