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3 years ago

{6, 7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 13,

Mathematics

1 answer:

Westkost [7]3 years ago

6 0

Answer:

Set 1 of Data:

Mean - 10

Median - 10

Set 2 of Data:

Mean - 16.5

Median - 10

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Find the midpoint of (6, 4), and (0,2).

Gnesinka [82]

Answer:

(3,3)

Step-by-step explanation:

Using the midpoint formula:

$m=\left( \dfrac{6+0}{2},\dfrac{4+2}{2} \right)=\left( \dfrac{6}{2},\dfrac{6}{2} \right)=(3,3)$

Hope this helps!

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3 years ago

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ale4655 [162]

Answer:

6=65

Step-by-step explanation:

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3 years ago

A moving company charges $0.60 per pound for a move from New York to Florida. A family estimates that their belongings weigh abo

m_a_m_a [10]

8000 x .60 = $48.00
This is because 4 tons = 8000 pounds.

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3 years ago

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Your parents have a realized income of $1,046.78/every 2 weeks. How much should the monthly housing expense budget be to stay wi

IgorLugansk [536]

Answer with explanation:

Income earned by my parents in every two weeks = $ 1046.78

In two weeks there are 14 days, so income earned by my parents in 1 day

$=\frac{1046.78}{14}\\\\74.77$

Income of my parents after a month, if there are 30 days in a month = $74.77 × 30= $ 2243.10

We, will use , $\frac{28}{36}$ rule here , which states that , 28% of monthly income of my parents, should be used for monthly housing expenses ,and 36% of gross monthly income should be used for debt purposes.

28% of $2243.78

$=\frac{28}{100}*2243.78\\\\=\frac{62825.84}{100}\\\\=628.2584$

= $ 628.26 (Approx)

So,my parents, monthly housing expense should not exceed ≥ $ 628.26

8 0

3 years ago

Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co

Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like $v_{1},v_{2},v_{3}, ...$ we call them linearly dependent if we have scalars $a_{1},a_{2},a_{3},...$ as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

$a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0$

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

$\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}$

So each column of this Matrix is a vector. So we can write them as:

$\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle$ Or

Now let's rewrite it as a system of equations:

$a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

$\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ -2 &4 &-4 &3 \\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &0 &9 &0\\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow R_{3}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &1 &-1 &4 \\ 0 &0 &9 &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )$ $\left\{\begin{matrix}1a_{1} &-2a_{2} &+2a_{3} &+3a_{4} &=0 \\ &1a_{2} &-1a_{3} &+4a_{4} &=0 \\ & & &9a_{4} &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0$

$S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}$

3 0

3 years ago