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Maurinko [17]
3 years ago
14

Find the range and the interquartile range of the data set represented by the box plot.

Mathematics
2 answers:
N76 [4]3 years ago
8 0
The interquartile range is 4 to 5.5 because where the lines fall that is the different quartile range
Fed [463]3 years ago
4 0

Answer: Range = 14

Interquartile range = 10

Step-by-step explanation:

From the given box plot, we can see the minimum value in data set = 4

The maximum value in data set = 18

First Quartile [starting point of box]Q_1=5.5

Third Quartile = [ending point of box]Q_3=15.5

Now, we know that the range of a data set is given by ;_

\text{Range = Maximum value - Minimum value}\\\\\Rightarrow\text{Range}=18-4=14

The interquartile range is given by :-

IQ=Q_3-Q_1\\\\\Rightarrow\ IQ=15.5-5.5=10

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If f(x) = 4 – x2 and g(x) = 6x, which expression is equivalent to (g – f)(3)?
Llana [10]
F(x) = -x² + 4
g(x) = 6x

(g - f)(3) = 6(3) - (-(3)² + 4)
(g - f)(3) = 18 - (-9 + 4)
(g - f)(3) = 18 - (-5)
(g - f)(3) = 18 + 5
(g - f)(3) = 23
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5 0
3 years ago
A garden table and a bench cost $600 combined. The cost of the garden table is three times the cost of the bench. What is the co
Sonja [21]
This is how I'd solve it:

bench = x           x+3x=600 
table = 3x               4x=600
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The bench costs <u>$150</u> and the table costs <u>$450</u>; combined, they cost $600, so it's correct. I hope this helps you!
3 0
3 years ago
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fredd [130]
The answer is in decimal formal in the file below

4 0
3 years ago
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What number makes the equation true? Enter the answer in the box
Reika [66]

Answer:

6

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4 0
3 years ago
During the first decade of the century, the population of a certain city was 145,380 in 2000 and 219,135 in 2010. Find the expon
Likurg_2 [28]

Answer:

The exponential growth function is P=145380e^{0.04103t}

Step-by-step explanation:

Given: The population of a certain city was 145,380 in 2000 and 219,135 in 2010.

To find: exponential growth function that models the growth of the city

Solution:

The exponential growth function is given by P=P_0 e^{kt}

Here, P denotes total population after time t

P_0 denotes initial population

k denotes rate of growth

t denotes time

As P(0)=145,380,

145380=P_0e^{0}\\145380=P_0\\P=145380e^{kt}

As P(10)=219135

219135=145380e^{10k}\\e^{10k}=\frac{219135}{145380}\\=1.507326\\k=\frac{1}{10}\ln (1.507326)\\=0.04103\\\Rightarrow P=145380e^{0.04103t}

8 0
3 years ago
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