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IrinaVladis [17]
3 years ago
7

in a 10 team league, each teams play every other team exactly twice. find the total number of games played in the league

Mathematics
1 answer:
LuckyWell [14K]3 years ago
7 0
Since it’s a 10 team league and each team plays with the other team twice then 10+10=20, thus there would be 20 games played
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3 square roots 7 multiplied by square root of 7 / 6 square roots of 7 raised to the 5
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Approximately 97.6149
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reflect (7,4) across the x-axis then reflect the result across the y-axis what are the coordinates of the final point
Andrews [41]

The coordinates of the final point after the two reflections is (-7, -4)

<h3>How to reflect a point across an axis?</h3>

For a point (x, y), a reflection across the x-axis gives the point (x, -y), while for a point (x, y) a reflection across the y-axis gives (-x, y).

In this case, we have the point (7, 4), first we reflect across the x-axis, so we get the point (-7, 4).

Now we reflect across the y-axis, so the sign of the y-component changes, and we get (-7, -4)

So the coordinates of the final point after the two reflections is (-7, -4).

If you want to learn more about reflections, you can read:

brainly.com/question/4289712

8 0
2 years ago
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
What is the measure of arc WXY
bonufazy [111]

Answer:

152°

Step-by-step explanation:

Let P be any point on tangent \overleftrightarrow{YZ} and WY is secant or chord of the \odot J .

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Sveta_85 [38]
146,304

You can get this by multiplying the amount of marbles in 1 box by how many boxes there are
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