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3 years ago

Which modified box plot represents the data set? 2, 8, 18, 7, 42, 12, 16, 14, 11, 4, 1, 10, 19

Mathematics

2 answers:

Lubov Fominskaja [6]3 years ago

7 0

Answer:

Step-by-step explanation:

Arrange the data 2, 8, 18, 7, 42, 12, 16, 14, 11, 4, 1, 10, 19 in ascending order:

1, 2, 4, 7, 8, 10, 11, 12, 14, 16, 18, 19, 42

The number 42 is an outlier of the data.

1, 2, 4, 7, 8, 10, 11, 12, 14, 16, 18, 19, 42

Write a five-number summary:

Min: 1

Q1: 5.5 (the average between the middle terms of first half of the data)

Med: 11 (the middle term of the data)

Q3: 17 (the average between the middle terms of second half of the data)

Max: 19

Outlier: 42

Only option D shows

Alenkasestr [34]3 years ago

5 0

Answer:

The last box plot

Step-by-step explanation:

A box plot is simply a graphical representation of the 5 number summary of a data set;

The minimum, first quartile, median, third quartile, and the maximum

in that order.

We can therefore arrange our given data set from the least to the greatest value;

1, 2, 4, 7, 8, 10, 11, 12, 14, 16, 18, 19, 42

The value 42 is an outlier since it is more extreme than the rest of the values.

From our arrangement of the data values above, we can note the following;

The minimum value is 1

The maximum value is 19

42 is an outlier, will be represented by a dark circle

The box-plot that matches the above is the last one and is thus the correct modified box-plot for the data

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An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

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A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

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Tom read 20 books, 12 magazines, and 18 newspaper articles last year.

neonofarm [45]

Answer: 24% of the books read are magazine

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Then, 12 of 50 =24%( 12÷50)× 100/1 = 24%

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Virty [35]

0 - (-6)

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egoroff_w [7]

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