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3 years ago

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Analyze the situation illustrated in the graphic. Which would typically be the result of osmosis so that the cell may maintain h

omeostasis? A. Water moves from the inside to the outside of the cell, causing the cell to swell. B. Water moves from the outside to the inside of the cell, causing the cell to swell. C. Water moves from the inside to the outside of the cell, causing the cell to shrivel. D. Water moves from the outside to the inside of the cell, causing the cell to shrivel.

1 answer:

bekas [8.4K]3 years ago

7 0

i think the answer is B

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4 years ago

Select ALL the correct answers.

tekilochka [14]

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$

$10th\ run = 9$

So:

$B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

Required

Select all true statements

(a) & (d) Median Comparisons

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$ $B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

$n = 9$ $n = 10$

Arrange the data:

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$Median = \frac{n + 1}{2}th$

$Median = \frac{9 + 1}{2}th$ $Median = \frac{10 + 1}{2}th$

$Median = \frac{10}{2}th$ $Median = \frac{11}{2}th$

$Median = 5th$ $Median = 5.5}th$ --- average of 5th and 6th

$Median = 2$ $Median = \frac{2+2}{2} = 2$

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$n = 9$ $n = 10$

First, calculate the lower quartile (Q1)

$Q_1 = \frac{n + 1}{4}th$ [Odd n] $Q_1 = \frac{n}{4}th$ [Even n]

$Q_1 = \frac{9 + 1}{4}th$ $Q_1 = \frac{10}{4}th$

$Q_1 = \frac{10}{4}th$ $Q_1 = 2.5$

$Q_1 = 2.5th$

This means that:

$Q_1 = 2nd + 0.5(3rd - 2nd)$ $Q_1 = 2nd + 0.5(3rd - 2nd)$

$Q_1 = 1 + 0.5(1- 1)$ $Q_1 = 1+ 0.5(1 - 1)$

$Q_1 = 1$ $Q_1 = 1$

Next, calculate the upper quartile (Q3)

$Q_3 = \frac{3}{4}(n + 1)th$ [Odd n] $Q_3 = \frac{3}{4}(n)th$ [Even n]

$Q_3 = \frac{3}{4}(9 + 1)th$ $Q_3 = \frac{30}{4}th$

$Q_3 = \frac{30}{4}th$ $Q_3 = 7.5th$

$Q_3 = 7.5th$

This means that:

$Q_3 = 7th + 0.5(8th- 7th)$ $Q_3 = 7th + 0.5(8th- 7th)$

$Q_3 = 2 + 0.5(3- 2)$ $Q_3 = 2+ 0.5(4 - 2)$

$Q_3 = 2.5$ $Q_3 = 3$

The interquartile range is $IQR = Q_3 - Q_1$

So, we have:

$IQR = 2.5 - 1$ $IQR = 3 - 1$

$IQR = 1.5$ $IQR =2$

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

8 0

3 years ago