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Luda [366]
2 years ago
10

Ms. Sutton recorded the word counts and scores of the top ten essays in a timed writing contest. The table shows her data. A 3-r

ow table with 10 columns. The first row is labeled essay rank with entries 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The second row is labeled essay score with entries 9. 7, 9. 5, 9. 5, 9. 0, 8. 9, 8. 9, 8. 5, 8. 4, 8. 2, 7. 0. The third row is labeled word count with entries 324, 352, 337, 318, 299, 304, 290, 291, 278, 250. How does the word count of an essay relate to its score in the contest? Word count tends to decrease as the score decreases. Word count tends to increase as the score decreases. Word count tends to remain constant as the score decreases. Word count has no apparent relationship to the score of the essay.
Mathematics
1 answer:
In-s [12.5K]2 years ago
6 0

Answer:

D. word count has no apparent relationship to the score of the essay

Step-by-step explanation:

ya its D ANYWAYSSSSSS

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33 is the answer :) Please give me the brainliest answer, and a rate and thanks.
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In its monthly report, the local animal shelter states that it currently has 24 dogs and 18 cats available for adoption. Eight o
andrezito [222]

Answer with Step-by-step explanation:

We are given that

Total dogs=24

Total cats=18

Total animals=24+18=42

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a.We have to find the probability that the pet is male, given that it is cat.

It means we have to find P(Male/cat)  

Conditional probability: P(A/B)=\frac{P(A\cap B)}{P(B)}

P(Cat)=\frac{18}{42}

P(male cat)=\frac{6}{42}

P(Male/cat)=\frac{\frac{6}{42}}{\frac{18}{42}}

P(male/cat)=\frac{6}{18}=0.33

b.P(female cat)=\frac{12}{42}

P(female)=\frac{28}{42}

P(Cat/Female)=\frac{\frac{12}{42}}{\frac{28}{42}}=\frac{12}{28}=0.43

c.P(Dog)=\frac{24}{42}

P(dog female)=\frac{16}{42}

P(female/dog)=\frac{\frac{16}{42}}{\frac{24}{42}}

P(female/dog)=\frac{16}{24}=0.67

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Answer:

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