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sveticcg [70]
3 years ago
8

Choose the correct answer below.

Mathematics
1 answer:
Bumek [7]3 years ago
5 0

Answer:

B. Descriptive statistics describes sets of data. Inferential statistics draws conclusions about the sets of data based on sampling.

Step-by-step explanation:

Let's explain on why the others are not,

A. Descriptive statistics is a characteristic or property of an individual experimental unit. Inferential statistics is the process used to assign numbers to variables of individual population units.

Descriptive statistics needs a set of data, not just an individual experimental unit, since it uses statistical methods like tendency, mean, median, mode, dispersion ranges, deviation, etc. That cannot be gathered from an individual set unit. In the same way, Inferential statistics, is about using a sample of data to infer, as the name says, behavior of larger groups, it doesn't assign numbers or variables, it studies behavior of samples, that being said, a variable might be "assigned" by testing an hypothesis, yet not from an unit, but from a set.

C. Descriptive statistics draws conclusions about the sets of data based on sampling. Inferential statistics summarizes the information revealed in data sets.

It is the complete oposite, Descriptive Statistics (DS), studies data to represent numerical analysis in charts and graphs, but not samples. While Inferential Statistics (IS), is about samples, and how they COULD represent a larger group.

D. Descriptive statistics are measurements that are recorded on a naturally occurring numerical scale. Inferential statistics are measurements that cannot be measured on a natural number​ scale; they can only be classified into one of a group of categories.

In IS, its not that the measurements cannot be measured on a natural number scale, it is that the size of the date is too large, so it needs to be broken down in order to be studied, it can have several categories but a smaller data set (thus a sample)

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The graph compares the weights in pounds of 100 dogs and cats that are brought in to a veterinarian's office.
GenaCL600 [577]
In case of dogs the value 10 is the minimum value. So all the values lie above 10. In total there were 100 dogs.
So for dogs, we can say number of dogs above the value of 10 pound are 100.

In case of Cats, 10 lies at the position of median. Median is the central value and 50% values lie above the median value. So number of cats with weight above 10 pound is 50.

Thus, we can conclude that there were 50 more dogs than the cats with weight over 10 pounds. So option C gives the correct answer.
8 0
3 years ago
Read 2 more answers
Solve the system of equations. Type in all points of intersection for the two functions and round to the nearest tenth if necess
Veseljchak [2.6K]

The points of intersection for the two functions and round to the nearest tenth are (0.5, 1.4) and (4.85, -0.3)

<h3>System of equations</h3>

Given the following function

f(x) = - 0.5x + 2

g(x) = x^3 - 5x^2 + 3

The point of intersection is the point where f(x) = g(x)

x^3 - 5x^2 + 3 = - 0.5x + 2

x^3 - 5x^2 + 3 + 0.5x - 2 = 0

x^3 - 5x^2 + 0.5x + 1 = 0

Factorize and determine the value of x

x = 0.53 and 4.85

If x = 0.53

f(0.53) = -0.5(0.53) + 2

f(0.53) = -0.265 + 2

f(0.53) = 1.375

If x = 4.85

f(4.85) = -0.5(4.85) + 2

f(4.85) = -2.425 + 2

f(4.85) = -0.245

Hence the points of intersection for the two functions and round to the nearest tenth are (0.5, 1.4) and (4.85, -0.3)

Learn more on functions here: brainly.com/question/10439235

#SPJ1

3 0
2 years ago
Margo borrows $1900, agreeing to pay it back with 4% annual interest after 9 months. How much interest will she pay?
lana66690 [7]

1900*4%

1900*0.04

76 (interest for a year)

76*3/4 (3/4 because 9 months is 3/4 of a year)

57 (9 months interest)

Margo will pay $57 in interest  



5 0
3 years ago
When originally purchased, a vehicle costing $23,000 had an estimated useful life of 8 years and an estimated salvage value of $
earnstyle [38]
The <u>correct answer</u> is:

$5,000.

Explanation:

When it is first purchased, the depreciation expense is calculated using the formula:

\text{Depreciation expense}=\frac{\text{Cost}-\text{Salvage Value}}{\text{Useful Life}}

The cost was $23,000; the salvage value was $3,000; and the useful life was 8 years:

\text{Depreciation Expense}=\frac{23000-3000}{8}=\frac{20000}{8}= \$2500

This means the value of the vehicle depreciates $2500 per year.

After 4 years, the vehicle would depreciate 2500(4) = $10,000.

This makes the new value $23000-$10000 = $13000.

Reevaluating the depreciation expense at this point, we use $13000 for the "cost" (current value), $3000 is still the salvage value, and now the total useful life was 6; we take 4 off of this, since it has already been 4 years:

\text{Depreciation Expense}=\frac{13000-3000}{6-4}=\frac{10000}{2}= \$5000

The depreciation expense in year 5 is $5,000.
6 0
3 years ago
Read 2 more answers
Please help (no links or you will reported)
goblinko [34]

Answer:

16 rolls of ribon

Step-by-step explanation:

C = Dπ

d = r x 2

d = 100 x 2

d = 200

C = 200 x 3.14

C = 628

628 / 40 = 15.7

she should buy 16 rolls

3 0
3 years ago
Read 2 more answers
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