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Shtirlitz [24]
4 years ago
13

A data set has values ranging from a low of 10 to a high of 52. what's wrong with using the class limits 10-19, 20-29, 30-39, 40

-49 for a frequency table?
Mathematics
1 answer:
tino4ka555 [31]4 years ago
7 0
When choosing an interval to use for a frequency table, the low value and the high value of the dat is considered so that the interval refrects a true data size for all the intervals.

Given that a<span> data set has values ranging from a low of 10 to a high of 52. Using the class limits of 10-19, 20-29, 30-39, 40-49 for a frequency table will make the last interval not to refrect the true size of the other intervals. i.e. the last interval will be 50 - 52 which has a size of 3, different from the size of 10 the other intervals have.</span>
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M=6 And Its Y-Intercept, b= -5
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Step-by-step explanation:

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B

Step-by-step explanation:

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6x² + 7x - 5

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3 0
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Read 2 more answers
A manufacturer claims that its drug test will detect steroid use (that is, show positive for an athlete who uses steroids) 95% o
amid [387]

Answer:

The probability tree is;

                            0.95        (+)

                   (S)

          0.1              0.05        (-)

[  P  ]

          0.9             0.15          (+)                                                        

                  (S_{no})    

                             0.85         (-)        

Step-by-step explanation:

Given the data in the question;

10% of the rugby team members use steroids

so Probability of using steroid; P( use steroid ) = 10% = 0.10

Probability of not using steroid; P( no steroid use ) = 1 - 0.10 = 0.90

Since the test show positive for an athlete who uses steroids, 95% of the time.

Probability of using steroids and testing positive = 95% = 0.95

Probability of using steroids and testing Negative = 1 - 0.95 = 0.05

Also from the test, 15% of all steroid-free individuals also test positive.

so

Probability of not using steroids and testing positive = 15% = 0.15

Probability of not using steroids and testing negative = 1 - 0.15 = 0.85

To set up the probability tree, Let;

(S) represent steroid use

(S_{no}) represent no steroid use

(+) represent test positive

(-) represent test negative

so we have;

                            0.95        (+)

                   (S)

          0.1              0.05        (-)

[  P  ]

          0.9             0.15          (+)                                                        

                  (S_{no})    

                             0.85         (-)        

7 0
3 years ago
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