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iVinArrow [24]
2 years ago
12

Descriptive statistics are generally used for which of the following? Check all that apply. To make general conclusions about po

pulations using sample data To describe independent variables only To describe dependent variables only To simplify and summarize data
Mathematics
1 answer:
Tom [10]2 years ago
8 0

Answer:

Correct answer is :

To simplify and summarize data

Step-by-step explanation:

Using sample data we can make inferences about a population using statistical methods such as linear regression. But descriptive statistics, as only describe general characteristics of a sample, can be not generalized to a larger population only by descriptive statistics.

Descriptive statistics (DS) can be used to describe dependent or dependent variables alike. If for example you want to predict the income of a person (dependent) in relation to age, education level ( independent) , descriptive statistics may be used to describe both type of variables the age of the sample or the levels of income.

DS are used basically to summarize data or to make it more simple to analyze.

You might be interested in
A point x is 34m due to east of a point Y . The bearing of a flagpole from x and Y are N18°W And N40°E respectively .
Murljashka [212]

Answer:

The distance of flag post from Y is 38.13 m

Step-by-step explanation:

Consider point Y at the intersection of both lines as shown below. Now the point X lies 34 meter from point Y in east direction.  

Now flag pole at point X lies at a bearing of N18°W. That is at point X from north, flag post makes an angle of 18° towards west.  

Similarly flag pole at point Y lies at a bearing of N40°E. That is at point Y from north, flag post makes an angle of 40° towards east.  

Consider ∆ AXY as right angle triangle. Therefore measure of angle FXY is,  

\angle AXY=\angle AXF+\angle FXY

\therefore 90 \degree=18\degree+\angle FXY

\angle FXY=72\degree

Consider ∆ BYX as right angle triangle. Therefore measure of angle FYX is,  

\angle BYX=\angle BYF+\angle FYX

\therefore 90 \degree=40\degree+\angle FYX

\angle FYX=50\degree

Refer attachment 1.  

From diagram consider the triangle FYX. To find the third angle that is ∠YFX can be calculated by using angle sum property of triangle.

∠YFX+∠FYX+∠FXY=180°

∠YFX+50°+72°=180°

∠YFX=58°

Refer attachment 2.  

Now the distance FY can be calculated using sin rule as follows,  

\frac{\sin X}{FY}=\frac{\sin F}{YX}=\frac{\sin Y}{FX}

Substituting the values,

\frac{\sin 72}{FY}=\frac{\sin 58}{34}=\frac{\sin 50}{FX}

Simplifying first two terms,

\frac{\sin 72}{FY}=\frac{\sin 58}{34}

Cross multiplying,

34\times\sin 72=FY\times \sin 58

\dfrac{34\times\sin 72}{\sin 58}=FY

FY=38.13 m

7 0
2 years ago
Rita makes and sells necklaces. The materials to make each necklace cost $2.75. Last week she made 8 necklaces and wants to make
Tresset [83]

Answer:

The error she made was that she was adding x and 2.75. She should subtract 2.75 from x.

Another mistake that she made was that she sold each for $7 assuming that she would make a profit of 78, but she should see each necklace for $12.5 so that she could make a profit of $78.

Step-by-step explanation:

The error she made was that she was adding x and 2.75.

She should write the equation as 8 (x - 2.75) = 78; as she spends $2.75 to make a necklace.

By using the correct equation: 8 (x - 2.75) = 78

=> 8x - 22 = 78

=> 8x = 78 + 22

=> 8x = 100

=> x = 100/8

=> x = 12.5

Another mistake that she made was that she sold each for $7 assuming that she would make a profit of 78, but she should see each necklace for $12.5 so that she could make a profit of $78.

Hope this helps you.

3 0
2 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
2 years ago
8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?
Lerok [7]

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

8 0
2 years ago
A group of entomologists has determined that the population of ladybugs at a local park can be modeled by the equation y = − 1.4
Oksanka [162]
<h3>Answer:</h3>

A) 177.568 thousand.

B) 125.836 thousand.

<h3>Step-by-step explanation:</h3>

In this question, it is asking you to use the equation to find the population of ladybugs in a certain year.

Equation we're going to use:

y = -1.437 x + 197.686

We know that the "x" variable represents the number of years since 2010, so that means our starting year is 2010.

Lets solve the question.

Question A:

We need to find the ladybug population is 2024.

2024 is 14 years after 2010, so our "x" variable will be replaced with 14.

Your equation should look like this:

y = -1.437 (14) + 197.686

Now, we solve.

y = -1.437 (14) + 197.686\\\\\text{Multiply -1.437 and 14}\\\\y=-20.118+197.686\\\\\text{Add}\\\\y=177.568

You should get 177.568

This means that the population of ladybugs in 2024 is 177.568 thousand.

Question B:

We need to find the ladybug population is 2060.

2060 is 50 years after 2010, so the "x" variable would be replaced with 50.

Your equation should look like this:

y = -1.437 (50) + 197.686

Now, we solve.

y = -1.437 (50) + 197.686\\\\\text{Multiply -1.437 and 50}\\\\y=-71.85+197.686\\\\\text{Add}\\\\y=125.836

This means that the population of ladybugs in 2060 would be 125.836 thousand.

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>
7 0
2 years ago
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