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1 year ago

Ms. Wall will roll a dice. What is the probability that she will roll an even number?

Mathematics

2 answers:

sveta [45]1 year ago

6 0

50% because of the sides of dice

Sergeu [11.5K]1 year ago

5 0

Answer:

50%

Step-by-step explanation:

half of the numbers on dice are even

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HELPP FOR BRAINLEST RIGHT NOW!!!!

Verdich [7]

Answer: Counter, 0, 0.

Step-by-step explanation:

Think about a clock. The hand of a clock goes clockwise. When you tighten something (righty tighty) you spin it clockwise. You can rotate an object, lets say a square, clockwise. You can also rotate it counterclockwise, in the other direction. Therefore, you can rotate an object clockwise and counterclockwise.

You can rotate a figure around any point, such as the center of the figure, the origin, or anywhere else. One common place to rotate a figure around, such as a square, is the origin. This is the center of the coordinate plane. This point is not up, down, left, or right at all from the center. This coordinate is (0, 0). Therefore, the next two blank spaces should both be filled with 0.

The blank spaces should look like this:

One direction is clockwise and the other is counterclockwise.

...

This can be any coordinate point such as the origin which is at ( 0 , 0 )

5 0

2 years ago

The scale on map is 1:500000

Arada [10]

Answer:

3,500 km

Step-by-step explanation:

Assuming that 1:500000 is meters you would have to convert to kilometers.

7 * 500000 / 1000 = 3500

5 0

3 years ago

Find the mean of the data. give your answer in decimal form

aleksandr82 [10.1K]

Mean = (1+1+2+3+3+3+4+5)/8 = 22/8 = 2.75

answer
mean = 2.75

8 0

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