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3 years ago

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

1 answer:

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

kodGreya [7K]

Approximately 97.6149

3 0

3 years ago

Read 2 more answers

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 $\odot J$ .

$\therefore m\angle WYZ + m\angle WYP = 180°\\(Straight \: line \: \angle 's) \\\therefore 104° + m\angle WYP = 180°\\\therefore m\angle WYP = 180°- 104° \\\red{\boxed {\bold {\therefore m\angle WYP = 76°}}} \\$

NOW, by tangent secant theorem:

$m\angle WYP =\frac{1}{2}\times m(\widehat{WXY}) \\\\76°=\frac{1}{2}\times m( \widehat{WXY}) \\\\76°\times 2 =m( \widehat{WXY}) \\\huge \purple {\boxed {\therefore m(\widehat{WXY}) = 152°}}$

8 0

2 years ago

a box contains 576 marbles of six different colors if there are 254 boxes how many marbles are in all the boxes

Sveta_85 [38]

146,304

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

3 0

2 years ago