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Anton [14]
3 years ago
7

Make a histogram, using a bin width of ten, to display the bowling scores for these 31 players: 87, 104, 79, 94, 117, 82, 72, 11

6, 105, 95, 88, 93, 109, 119, 75, 103, 112, 97, 73, 85, 91, 86, 102, 99, 106, 84, 98, 83, 81, 96, 92.
Mathematics
1 answer:
slavikrds [6]3 years ago
7 0

Step-by-step explanation:

Using R, I used the following code to create a histogram:

bowling.scores <- c(87, 104, 79, 94, 117, 82, 72, 116, 105, 95, 88, 93, 109, 119, 75,  

                   103, 112, 97, 73, 85, 91, 86, 102, 99, 106, 84, 98, 83, 81, 96, 92)

data.frame(bowling.scores)

ggplot(data.frame(bowling.scores), aes(x=bowling.scores)) +  

 xlim(c(70, 120)) +

 scale_y_continuous(breaks = seq(0, 10, by=1), "Frequency") +

 geom_histogram(breaks=seq(70, 120, by=10), color="black", fill="grey60") +

 labs(title="Histogram of Bowling Scores", x="Bowling Scores", y="Frequency")

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What is 7 (1+10p) + 8 (1 + 6p)
Komok [63]

Answer:

58p+15

Step-by-step explanation:

First, we want to get rid of the parentheses in the equation. To do that, let's start with 7(1+10p). We multiply 7 by each item in the parenthesis, and get 7+70p. Keep that little thing in mind, we will use it later. Next, same process with the other part, and we get 8+48p. We multiplied both parts out, so now we just add like terms. 7 and 8 don't have letters, so we add them to get 15. 10p and 48p do have letters, so we add those together and get 58p. We can't add that to the other numbers, because 58p is 58 times p, and 15 doesn't have that p so it wouldn't work without knowing p. What we are left with, by simplifying the problem, is 58p+15, or 15+58p. The order doesn't matter, as long as you have that answer.

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3 years ago
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Answer:

C) 118.8; 28.8

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iogann1982 [59]
<h3>Answers:</h3>
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