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3 years ago

Which option has the same median and mode? 1,2,1,3,4 3,4,3,5,6 6,6,7,9,8 9,9,9,2,2

Mathematics

2 answers:

klio [65]3 years ago

7 0

Answer:

Step-by-step explanation:

sweet-ann [11.9K]3 years ago

3 0

Answer:

The last option → 9,9,9,2,2

Step-by-step explanation:

The median or middle number is 9.

The number that occurs the most often is also 9.

You might be interested in

If y = 2x and 3x - 3y = 2, then x =

faltersainse [42]

Answer:

The value of x is equal to -2/3.

Step-by-step explanation:

In the problem it says that y equals 2x, and that 3x - 3y = 2.

The first step is to substitute 2x into y.

3x - 3(2x) = 2.

The next step is to use distributive property.

-3(2x) = -6x.

Now we need to add like terms.

3x - 6x = -3x.

Which gives us the equation -3x = 2.

The final step is to divide on both sides to get the value of x.

-3x/-3 = 2/-3.

x = -2/3.

So as you can see, x is equal to -2/3.

We can also check to make sure by redoing the problem, but substituting the value of x.

3(-2/3) - 3(2 * -2/3) = 2.

-2 - 3 * -4/3 = 2.

-2 + 4 = 2.

2 = 2.

The value of x is indeed -2/3.

5 0

3 years ago

Calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)

Monica [59]

Answer:

$\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}$

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices $\mathsf{A(x_A, y_A)}$ , $\mathsf{B(x_B, y_B)}$ and $\mathsf{C(x_C, y_C)}$ , have a look at the representation in the cartesian plan.

From this representation we can say that the area (A) of a triangle through the knowledge of analytical geometry is given by the determinant of the vertices divided by two, mathematically,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} \mathsf{x_A} & \mathsf{y_A }& 1 \\ \mathsf{x_B} & \mathsf{ y_B} & 1 \\ \mathsf{ x_C} & \mathsf{ y_C} & 1 \end{array} \right|}{2}$

So, applying this knowledge we're going to have,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2 & -3 & 1 \end{array} \right|}{2}$

$\mathsf{A} \triangle = \dfrac{1}{2}\left[ \left.\begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2& -3 & 1 \end{array} \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]$