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Mamont248 [21]
3 years ago
7

The table represents the number of cheese crackers in the lunchboxes of 9 boys and 9 girls. By looking at the table, does it app

ear that the degree of variability for the boys' data is greater, less, or the same as the girls' data? Compute the interquartile range of each data set. Using the interquartile range, compare the degree of variability between the data sets. Explain how the comparison supports your first answer.
A) no; Boys: IQR: 4.5; Girls IQR: 3; The degree of variability for both sets is about the same.
B) no; Boys: IQR: 3; Girls IQR: 4.5; The degree of variability for both sets is about the same.
C) yes; Boys: IQR: 8; Girls IQR: 5; The degree of variability is greater for the boys than for the girls.
D) yes; Boys: IQR:10; Girls IQR:16; The degree of variability for the boys is less than for the boys.
Mathematics
2 answers:
mariarad [96]3 years ago
7 0

Answer:A)

Step-by-step explanation:

I got it wrong for you, lol. sorry im so late but i was looking for this awnser two.

Westkost [7]3 years ago
5 0
CAN SOMEONE PLEASE ANSWER IT RIGHT NOW I NEED IT!!
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Answer:

(-6,4)

Step-by-step explanation:

You started at (-4,2)  The first number in the ordered pair moves the number left and right and the second number moved the point up and down.

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Graph the following three equations on your own device and answer the questions below.
Alina [70]

Due to length restrictions, we kindly invite to read the explanation of this question for further details on functions.

<h3>What are the characteristics of each of the three functions?</h3>

a) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(10) = 5 · 10 + 7

f(10) = 57

Case 2

f(10) = 10² + 6

f(10) = 106

Case 3

f(10) = 2¹⁰ + 3

f(10) = 1027

b) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(100) = 5 · 100 + 7

f(100) = 507

Case 2

f(100) = 100² + 6

f(100) = 10006

Case 3

f(100) = 2¹⁰⁰ + 3

f(100) = 1.267 × 10³⁰ + 3

c) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(1000) = 5 · 1000 + 7

f(1000) = 5007

Case 2

f(1000) = 1000² + 6

f(1000) = 1000006

Case 3

f(1000) = 2¹⁰⁰⁰ + 3

f(1000) = (1.268 × 10³⁰)¹⁰ + 3

f(1000) = 10.744 × 10³⁰⁰ + 3

f(1000) = 1.074 × 10³⁰¹ + 3

e) The <em>third</em> function increases the fastest.

f) In this part we need to compare the <em>third</em> function with respect to the <em>first</em> and <em>second</em> functions:

5 · x + 7 = 2ˣ + 3

2ˣ - 5 · x   = 4

The solutions of the equation are x = - 0.675 and x = 4.81. The function will exceed the other <em>first</em> function at x = 4.81.

x² + 6 = 2ˣ + 3

2ˣ - x² = 3

The solution of the equation is x = 4.588. The function will exceed the other <em>second</em> function at x = 4.588.

g) Yes, <em>exponential</em> functions with bases greater than 1 will surpass <em>polynomic</em> function at any point x such that x > 0.

h) The domain represents the set of x-values of a function and the range represents the set of y-values of a function. Then, the domain and range of each function is:

Case 1

Domain - All <em>real</em> numbers.

Range - All <em>real</em> numbers

Case 2

Domain - All <em>real</em> numbers.

Range - [6, +∞)

Case 3

Domain - All <em>real</em> numbers.

Range - (3, +∞)

To learn more on functions: brainly.com/question/12431044

#SPJ1

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1 year ago
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