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Travka [436]
3 years ago
5

An older adult client presents with noted burning pain along the thoracic region with a tingling sensation and extreme sensitivi

ty of the skin to touch, but there is no rash present at this time. Which skin problem will the health care provider interpret for this client?
Biology
1 answer:
Westkost [7]3 years ago
5 0
The skin care will use the period prevent pain so the eggs don’t dye well the kids also the pain
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Would an enzyme work in a different environment that is used to?
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Enzymes work best within specific temperature and pH ranges, and sub-optimal conditions can cause an enzyme to lose its ability to bind to a substrate. Temperature: Raising temperature generally speeds up a reaction, and lowering temperature slows down a reaction.

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Select ALL the correct answers.
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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

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