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3 years ago

How can you recognize and interpret associations in two-way tables?

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

Answer:

Sometimes the best way to tell whether two variables are associated is to ask yourself whether they are not associated. Think backward. In a two-way frequency table, if the relative frequencies for one variable are the same (or close) for all categories of another variable, there is no (or little) associa

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Assume that 12 jurors are randomly selected from a population in which 73% of the people are Mexican-Americans. Refer to the pro

kumpel [21]

Answer:

I'm not sure if i'm correct but i'm guessing no

Step-by-step explanation:

6 0

3 years ago

Please help me thnk you

kramer

For this question, we use the Pythagorean theorem.

$a^2+b^2=c^2$

We know the lengths of the legs, a and b.
$a=3$
$b=4$

Now, let's solve for the hypotenuse, c.

$3^2+4^2=c^2$
$25=c^2$

Take the positive square root.

$\boxed{c=5}$

Hope this helps! :)

8 0

4 years ago

Read 2 more answers

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

1 year ago

#2) The Ramy family went

suter [353]

If there were only sandwiches bought, there would be 15, if there were only salads bought there would be 20. There’s probably more than one possibility, but that’s the best I can do, considering how long it’s been since I’ve done this type of problem. I’m all the way to learning determinants, so I really have to think to remember these. Sorry in advance if this doesn’t help at all, I’m not the best at this, but I can still try.

5 0

2 years ago

Y = (x) = (1/16)^x<br> Find f(x) when x = (1/4)

Naddik [55]

Answer:

1/2

Step-by-step explanation:

(1/16)^x

Let x = 1/4

(1/16)^ 1/4

Rewriting 16 as 2^4

(1/2^4)^ 1/4

We know that 1 / a^b = a^-b

(2 ^ -4)^ 1/4

We know that a^b^c = a^(b*c)

2^(-4*1/4)

2^-1

We know that a^-b = 1/ a^b

2^-1 = 1/2^1 = 1/2

7 0

3 years ago

Read 2 more answers