An avid golf fan recorded the scores of 75 golfers on the ninth hole of the Cottonwood Golf Course as follows :

Mathematics

1 answer:

Vadim26 [7]3 years ago

3 0

Answer:

The correct answer is 3.77 strokes as the average score for the ninth hole of the Cottonwood Golf Course.

Step-by-step explanation:

1. To find the average score, we need to know how many strokes the 75 golfers needed to play this ninth hole. For doing that, we will do the following addition, using the frequency data given:

1 x 1 = 1

1 x 2 = 2

32 x 3 = 96

27 x 4 = 108

8 x 5 = 40

6 x 6 = 36

1 + 2 + 96 + 108 + 40 + 36

283

This means that all the 75 golfers scored 283 strokes for playing the ninth hole

2. Now that we know that the total number of strokes was 283, we will do a division for find the average score, using also the number of golfers. For doing it, we will do the following division:

Total number of strokes/ total number of players

283/75 = 3.77

The average score is 3.77. It means that the 75 golfers in average hit 3.77 strokes to play the ninth hole.

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Mary took a math test and she got four questions incorrect and 21 correct. What percent did answer correctly?

AnnZ [28]

Answer:

84%

Step-by-step explanation:

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6 0

3 years ago

Please help! 30 points!

Rashid [163]

Answer: $64.2^{\circ}$

Step-by-step explanation:

According to the cosine rule

$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Here, $a=83, b=111, c=165,\ C=180^{\circ}-\theta$

Putting values

$\Rightarrow \cos (180^{\circ}-\theta)=\dfrac{83^2+111^2-165^2}{2\times 83\times 111}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{6889+12321-27225}{18426}\\\\\Rightarrow \cos(180^{\circ}-\theta)=\dfrac{-8015}{18426}=-0.4349\\\\\Rightarrow 180^{\circ}-\theta=\cos^{-1}(-0.4349)=115.77^{\circ}\\\Rightarrow \theta=64.23\approx 64.2^{\circ}$