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Alla [95]
2 years ago
9

Swati recorded this data set, which contains an outlier. 163, 97, 184, 199, 169, 175 What numbers represent the lower and upper

quartiles? lower quartile: 163 upper quartile: 184 lower quartile: 163 upper quartile: 172 lower quartile: 172 upper quartile: 184 lower quartile: 184 upper quartile: 163
Mathematics
2 answers:
Damm [24]2 years ago
8 0

Answer:

lower quartile  163

upper quartile is equal to 184.

arsen [322]2 years ago
6 0
To determine the answers to this item, we are to arranged the given data points in increasing order. The arrangement would then be,

         97,163, 169, 175, 184, 199

The lower quartile is the median of the lower half of the data points. In this case, 97, 163, 169. The median is equal to 163

The upper quartile is the median of the upper half of the data points. In this case, 175, 184, 199. The median is equal to 184.

Hence, the aswers to this item is that lower quartile is equal to 163 and upper quartile is equal to 184. The first option. 
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If the parent function is f(x) = x3, which transformed function is shown in the graph? g(x) = (x − 3)3 , g(x) = (x + 3)3 , g(x)
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A. cause  i did it on the calculator so i know its right
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Which is the best description of the graph of the function f(x) = 60(One-third)x? The graph has an initial value of 20, and each
dezoksy [38]

Answer:

The graph has an initial value of 60, and each successive term is determined by multiplying by One-third.

Step-by-step explanation:

Which is the best description of the graph of the function f(x) = 60(One-third)x?

A. The graph has an initial value of 20, and each successive term is determined by subtracting One-third.

B. The graph has an initial value of 20, and each successive term is determined by multiplying by One-third.

C. The graph has an initial value of 60, and each successive term is determined by subtracting One-third.

D. The graph has an initial value of 60, and each successive term is determined by multiplying by One-third.

Given:

f(x) = 60 (1/3)^x         see graph attached.

Examining the graph, we notice that

1. The graph has an initial value of 60, at point (0,60)

2. The graph decreases as x increases.  Each successive term is determined by multiplying by 1/3, namely in the sequence

{(0,60), (1,20), (2, 20/3), (3, 20/9), ...}

This means the choice is D (or the fourth one).

4 0
2 years ago
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The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the n
MArishka [77]

9514 1404 393

Answer:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

  142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

__

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

  (0.4/1200)(N)(1200-N) -50 = 0

  N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient

  N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square

  (N -600)^2 = 210,000 . . . . . simplify

  N = 600 + √210,000 ≈ 1058

This steady-state number of fish is ...

  1200 - 1058 = 142 . . . . below the original carrying capacity

8 0
2 years ago
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