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atroni [7]
2 years ago
9

Social support and stroke-induced aphasia. Hilari and Northcott (2006) used a Social Support Survey (SSS) to gauge how well-supp

orted individuals suffering from stroke-induced aphasia (a language disorder) felt more than one year following the stroke. They reported that “in terms of social support, the SSS scores were negatively skewed with a mean of 3.69, suggesting that participants felt overall well supported” (Hilari & Northcott, 2006, p. 17). Based on their findings, what additional measure of central tendency would be appropriate to report with these data?
Explain.
Mathematics
1 answer:
Nana76 [90]2 years ago
3 0

Answer:

A third metric to consider is the median. The reason for this is that all of the data is negatively skewed, and using mean would result in a reduction.

Step-by-step Explanation:

Okay, So for this problem, we were asked to make a relative frequency distribution based office of this clinical trial for this drug and its adverse side effects. Okay, so, um, that's look at the data here. We'll make a relative frequency distribution as a chart with two sides over here, we will write thes side effects. So I defects. Over here, we will write the frequency of those side effects. Okay, so we have headaches. Eggs. There was hypertension, a pretension, upper respiratory infection. A margarita, as you are. A per respiratory nasal Farron Ghitis. It's just right neighs o on then. Lastly, there was diarrhea. Okay, So the frequencies for these, we were told, were 57 in 21 60 51 and 53. Okay, so that is the frequency distribution right there. That's the first part of our answer. Okay, Second part of our answer says, are any of these, um, more common than the other? That's has to use relative frequency. Okay, So for relative frequency, relative frequency, it's just equal to the individual frequency f over the some summation symbol right there of the frequencies. Okay. And from that we'll get a percentage if we times it by 100%. Yeah. Cool. So, um, what is the sum of all these frequencies? Some of all the frequencies. It's just equal to 57 plus 21 plus 60 plus 51 plus 53. And that is just 242. So if I wanted to get the relative frequency for headaches, um, which I will do right now, let me switch colors to do that, do that and read on. So the relative frequency for headaches it's equal to 57 over 2 42 times, 100%. And when I do that, I get about 23%. So 23% Okay, I do it for the next one. I'll dio 21 divided by 2 42 times 100%. This is for hypertension. I get approximately 9%. Okay, So frequency of hypertension is 9%. It's a lot less than the headaches. Let's do the rest. So 60 divided by 2 42 times 100% equal to about 25%. And then the 51 divided by 2. 42 times 100% approximately 21% and then last 1 53 developed by 2 42 times 100% approximately 22%. Okay, fill out our chart here. So well, we have 25% for a respiratory. 51 21% for needs affair and itis. And then lastly for diarrhea, 22%. Okay, So the question asked us, um if any one of these air more common than the other Not really. There's less for hypertension, but they're all about more or less the same, so there is no difference or no prevalence of adverse side effects.

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Find the value of x. If necessary, you may learn with the markings on a figure indicate.
love history [14]

Answer:

112

Step-by-step explanation:

6 0
3 years ago
rectangle A has a length of 24 meters and a width of 20 meters. rectangle B has the same area as rectangle A. the width of recta
blagie [28]
To solve this, first we'll find the area of the rectangle A,

Area=length × width
?=24m×20m
480m=24m×20m
480m squared=area of the rectangle A

now we'll find the width of rectangle B,
"the width of rectangle B is 12 meters less than the width of rectangle A",
20m-12m= 8m
8m=width of rectangle B

finally we'll find the length of rectangle B,
area of the rectangle B= 480msquared
width= 8m
length=? (to find this divide the area by the width)
480÷8=60m
length of the rectangle B=60m



7 0
4 years ago
Here you go amiragumbs2
Simora [160]
Umm , what is this for? do you need help on anything or? lol
5 0
3 years ago
I made a square frame for my favorite bird picture from four wooden pieces. Each piece is a rectangle with a perimeter of 24 inc
kupik [55]

Answer:

The perimeter of the picture frame is 38.4 in.

The area of the picture frame is 92.6 in.².

Step-by-step explanation:

The given information are;

Perimeter of side piece of picture frame = 24 inches

Length of side piece = L

Width of side piece = W

Perimeter of side piece = 2 × (L + W) = 24

∴ L + W = 24/2 = 12 inches

Dimension of picture frame = Square frame with side length s

Number of side piece in picture frame = 4

Given that the length L > the width W, we have

Side length of wooden frame = L

Also, where the side piece are placed side by side, we have;

Side length of wooden frame = 4 × W

Therefore;

4 × W = L

Which gives

L + W = 12 inches

4 × W + W = 12 inches

W×(4 + 1) = 5·W = 12 inches

W = 12/5 = 2.4 inches

L = 4 × W = 4 × 12/5

L = 48/5 = 9.6 inches  Side length of wooden frame, s

The perimeter of the picture frame = 4 × s = 4 × 9.6 = 38.4 in.

The area of the picture frame = s × s = 9.6 × 9.6 = 92.6 in.².

7 0
3 years ago
Which equation could represent the relationship shown in the scatter plot?
KiRa [710]
D? D just makes the most sense i suppose 

3 0
3 years ago
Read 2 more answers
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