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vredina [299]
3 years ago
11

A professional basketball player participated in 1,558 games in his career. He randomly chose eight games to determine the numbe

r of points he made per game, as shown below. 27, 19, 29, 19, 25, 27, 25, 27 If the sample was representative of his entire career, what was the mode of the number of points per game?
Mathematics
1 answer:
DENIUS [597]3 years ago
4 0

Answer:

27 is the mode of the number of points per game.  

Step-by-step explanation:

We are given the following in the question:

Total number of games = 1558

Sample size = 8

Sample of points made per game:

27, 19, 29, 19, 25, 27, 25, 27

Sorted data:

19, 19, 25, 25, 27, 27, 27, 29

Mode:

  • It is the most frequent observation in the sample.

Since 27 repeats itself three times that is highest repeating number, the mode of the data is 27.

27 is the mode of the number of points per game.

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Solve for x. Enter your answer as an integer or fraction. -9 -4x=7
jeka57 [31]

-9 -4x = 7

add 9 to both sides in order to isolate the x

-4x = 16

divide by -4

x = -4

6 0
3 years ago
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Factor yz + 6 + y +6z.<br> z + 1)(y + 6)<br> (z - 1)(y + 6)<br> (z + 6)(y - 1)
Sedbober [7]
The answer is the first one: (z+1)(y+6).

Explanation:

yz + 6 + y + 6z

= yz + y + 6 + 6z

= y(z+1) + 6(1+z)

= (z+1)(y+6)

For (z+1), you can also write (1+z), it doesn’t matter.

Hope it helps!

8 0
3 years ago
Use the following graph to solve the equation 3n + 7 = 52
Tanya [424]

Answer:

3n + 7 = 52 \\ 3n = 52 - 7 \\ 3n = 45 \\ n =  \frac{45}{3}  \\ n = 15

Graph is not inserted.

8 0
2 years ago
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part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.​
Nat2105 [25]

Answer:  \bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}

<u>Step-by-step explanation:</u>

Pythagorean Theorem is: a² + b² = c²   , <em>where "c" is the hypotenuse</em>

<em />

9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}

Note: 4² + (8√2)² = hypotenuse²   →   hypotenuse = 12

10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}

Note: 12² + opposite² = 20²   →   opposite = 16

11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}

Note: adjacent² + 5² = 6²   →   adjacent = √11

12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}

Note: adjacent² + 7² = (13√2)²   →   adjacent = 17

5 0
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The graph shown can be used to solve which of these systems of equations?
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Answer: It is B

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