#every one when im in public be like

Select each of the following experiments that are binomial experiments:

Vinvika [58]

Answer:

a) No.

b) Yes.

c) Yes.

Step-by-step explanation:

a) No.

As being without replacement, the probabilities of each color in each draw change depending on the previous draws.

This is best modeled by an hypergeometric distribution.

b) Yes.

As being with replacement, the probabilities for each color is constant.

Also, there are only two colors, so the "success", with probability p, can be associated with the color red, and the "failure", with probability (1-p), with the color blue, for example.

(With more than two colors, it should be "red" and "not red", allowing only two possibilities).

c) Yes.

The answer is binary (Yes or No) and the probabilities are constant, so it can be represented as a binomial experiment.

8 0

3 years ago

What does chemical erosion do

matrenka [14]

It causes the breakdown and decay to the rocks. It occurs when the weather water is slightly acidic.

5 0

3 years ago

Jackson bought 5 ounces of raisins for $4, how many ounces cna he buy for $1.

Zina [86]

He can buy 1 1\4 ounce for a dollar or 1.25 ounces or depending on your answer choice 1 ounce. All those answer are right

7 0

3 years ago

Read 2 more answers

Explain why a graph that fails the vertical-line test does not represent a function. Be sure to use the definition of a function

ArbitrLikvidat [17]

Explanation:

A function is a relationship where any one x-value/input only has <u>one </u>corresponding y-value/output. (note: a y-value can have multiple x-values).

> This can be called "assigning one y-value to every x element".

The vertical line test places a line that would connect all y-values of an x-value (that is, if it were to have multiple y-values). If multiple points can be found along the vertical line, it is, therefore, by the definition of a function, not a function. (Because an x-value will have more than one y-value).

So, a graph that fails the vertical-line test does not represent a function because an x-value will correspond with more than one y-value.

hope this helps!!

4 0

2 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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6 0

1 year ago