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babymother [125]
3 years ago
14

PLEASE HELP WILL GIVE BRAINLIEST!!!!These box plots show the basketball scores for two teams

Mathematics
2 answers:
aev [14]3 years ago
6 0

Answer:

B: The interquartile range (IQR) for the Bulldogs, 20 is LESS than the IQR for the tigers 25.

lyudmila [28]3 years ago
4 0

Answer:

B. The Interquartile Range (IQR) for the Bulldogs, 20, is less than the IQR for the Tigers, 25

Step-by-step explanation:

The question is asking for a comparison of spread. There are two methods for  measuring spread, that are standard deviation and the Interquartile Range. For a skewed box plot like this one you would use the Interquartile Range. The 5 numbers on a box plot are part of the 5 number summary. From left to right the numbers above the box plot are the minimum, the first quartile (Q1), the median (Q2), the third quartile (Q3), and the maximum. To find IQR you just subtract the third quartile (Q3) from the first (Q1). In this case, for the Bulldogs the IQR= 90-70+=20 and for the Tigers IQR=85-60=25.

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Answer:

Square.

Step-by-step explanation:

given J(-4,2) || K(0,3) || L(1,-1) || M(-3,-2)

given below the visual answer.

OR

you can find the distance of coordinate to coordinate:

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√(-4-0)² + (2-3)²

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Distance of LM

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Distance of MJ

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2 years ago
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Step-by-step explanation:

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3 years ago
Comparing Representations of Modeled Relationships
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A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the individuals who actually have the
Ne4ueva [31]

Answer:

0.2177 = 21.77% conditional probability that she does, in fact, have the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

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Probability of a positive test:

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P(A) = 0.90*0.03 + 0.1*0.97 = 0.124

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0.2177 = 21.77% conditional probability that she does, in fact, have the disease

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