For a fixed total number of boys and girls, there is only one possible answer. The boys to girls ratio is 3:2, which means 3/5 are boys and 2/5 are girls. If the total number of boys and girls is N, then
number of boys = (3/5)N
number of girls = (2/5)N
Obviously, N must be a multiple of 5, since we cannot have "fractional" boys or girls.
Answer:
1610
Step-by-step explanation:
23 x 14 / 0.2 = 1610
Hello!
As you can see, there are 9 total activities. As you can see, 6 of them burn more than 400 calories. This gives us the probability 6/9.
Therefore, our answer is C) 6 out of 9
I hope this helps!
Two numbers = 37
Larger number is x, so the smaller number is 37 - x:
x + (37 - x) = 37
“The difference” is subtraction, so we are subtracting the smaller and larger number.
“Nine more than twice the larger number” looks like 9 + 2x, and “the sum of the smaller number and three” looks like (37 - x) + 3. Because this is subtraction, we would write (9 + 2x) - (37 - x) + 3. First, distribute the negative sign:
-(37 - x) = -37 + x
We then simplify:
9 + 2x - 37 + x + 3 (subtract 3)
9 + 2x - 34 + x
-25 + 2x + x (subtract 8 and 34 then add 2x and x)
-25 + 3x
Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
In conclusion, AX=B is:
