5,3,1,-1, common difference or common ratio

Assume that 2 cards are drawn from a standard 52-card deck. Find the following probabilities.

topjm [15]

Answer:

Step-by-step explanation:

WITHOUT replacement of first card drawn:

P(a 10 is drawn) = 13/52 = 1/4

P(the next draw is a 10) = 12/52 = 3/13

P(drawing two 10s without replacement of the first draw) = (1/4)(3/13) = 3/52

WITH replacement of first card:

P(two 10s are drawn) = P(first card is a 10)*P(first card is a 10) = (4/13)(4/13) =

16/169

7 0

3 years ago

What is the row echelon form of this matrix?

USPshnik [31]

The row echelon form of the matrix is presented as follows;

$\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}$

<h3>What is the row echelon form of a matrix?</h3>

The row echelon form of a matrix has the rows consisting entirely of zeros at the bottom, and the first entry of a row that is not entirely zero is a one.

The given matrix is presented as follows;

$\begin{bmatrix}-3 &6 &15 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}$

The conditions of a matrix in the row echelon form are as follows;

There are row having nonzero entries above the zero rows.
The first nonzero entry in a nonzero row is a one.
The location of the leading one in a nonzero row is to the left of the leading one in the next lower rows.

Dividing Row 1 by -3 gives:

$\begin{bmatrix}1 &-2 &-5 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}$

Multiplying Row 1 by 2 and subtracting the result from Row 2 gives;

$\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 1&0 &-1 \\\end{bmatrix}$

Subtracting Row 1 from Row 3 gives;

$\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&2 &4 \\\end{bmatrix}$

Adding Row 2 to Row 3 gives;

$\begin{bmatrix}1 &-2 &-5 \\ 0& -2 & 14\\ 0&0 &18 \\\end{bmatrix}$

Dividing Row 2 by -2, and Row 3 by 18 gives;

$\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}$

The above matrix is in the row echelon form

Learn more about the row echelon form here:

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8 0

1 year ago

The average cost of tuition, room and board at small private liberal arts colleges is reported to be $8,500 per term, but a fina

Vlada [557]

Answer:

The test statistic for this test is 3.82.

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 8500$

The alternate hypotesis is:

$H_{1} > 8500$

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation of the population and n is the size of the sample.