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4 years ago

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the area of the hartstein's kitchen is 182 square feet. this is 20% of the area of the first floor of their house. let f represe

nt the area of the first floor. write an equation to represent the situation

1 answer:

Sav [38]4 years ago

4 0

F=182/.8 should be the correct equation to use in this problem since your trying to find the remaining 80% left of the floor

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(10-3)*2+(5-14/2 help

Dafna11 [192]

2(10-3) distribute = 2*10 is 20 and 2*3 is 6
20-6+(5-14/2)
now 20-6 is 14
14+(5-14/2)
now 14/2 is 7
14+(5-7)
now 5-7 is -2
14+-2
is 12
the answer is 12
but if you want to you can do 12/3 to get the greatest common factor
which is 4
hope i help you :P

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3 years ago

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faltersainse [42]

Answer:

a

Step-by-step explanation:

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3 years ago

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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3 years ago

Complete the rule:<br> positive + negative -

aivan3 [116]

Answer: If you have a positive and a negative, there is one dash left over, and the answer is negative.

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What is the r-value of the following data to three decimal places

Liula [17]

Answer:

option B

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Step-by-step explanation:

we know that

The<u> correlation coefficient</u> r measures the direction and strength of a linear relationship.The value of r is always between $+1$ and $-1$

Using a Excel tool (Correl function)

see the attached table

The value of r is $-0.953$

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3 years ago

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