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2 years ago

What is the median of Variable A? Express answer to one decimal place. _____ Fill The Blank

Mathematics

1 answer:

skelet666 [1.2K]2 years ago

8 0

The median is the number that is in the middle of the set of numbers in order from least to greatest. So in this case we have...1,2,4,4,5,6,7,9 and you cross out one from the left one from the right over and over until you are left with one or two numbers
LOOK AT ATTACHMENT!!!

Now that we have narrowed it down to 4 and 5, we need to add them together (9) then divide it by 2 which would give you the answer of 4.5

So the median of Variable A is 4.5

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Help with triangles 8th grade

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2 years ago

An investment firm invested in two companies last year. They invested $11,000 in Company A and made a profit of 24% . They inves

Tema [17]

First off, "whatever%" of "anything" is just (whatever/100) * anything

part A)

so... firm A got 11,000 invested, turned a profit of 24%, how much is 24% of 11000? well (24/100) * 11000

firm B got 14,000 invested, and returned 15% in profits, how much is 15% of 14000? (15/100) * 14000

part B)

for the amounts above, we get 2640 and 2100 respectively

so, the total profit "amount" is 2640 + 2100 or 4740

the total investement was 11000+14000 = 25000

if 25000 is the 100%, how much is 4740 in percentage?

$\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
25000&100\\
4740&x
\end{array}\implies \cfrac{25000}{4740}=\cfrac{100}{x}$

solve for "x"

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2 years ago

Read 2 more answers

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

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2 years ago

True or false? it's not possible to build a triangle with side lengths of 3 3 and 9

Anit [1.1K]

True

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Vilka [71]

If you are looking for the value of N, multiply both sides by 4. The value of N is 4900.

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3 years ago