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Akimi4 [234]
2 years ago
8

Predictor variables are:___.

Mathematics
1 answer:
timofeeve [1]2 years ago
7 0

The word Predictor variables are known to be used in regression to predict values of the variable of interest,  Variables whose values can be easily determined and are related to the variable of interest.

<h3>What is a predictor variable?</h3>

Predictor variable is known to be a kind of variable where the independent variable is often used in regression analyses.

The predictor variable helps one with the information on a linkage of the dependent variable in terms of a particular outcome. In fundamental level, predictor variables are regarded as variables that are related with a specific outcomes.

Learn more about Predictor variables from

brainly.com/question/7781679

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Harold bought 5 apples for $ 1.30. What is the unit price?
Firdavs [7]
What is unit price?- unit price is the cost of one item(which in this case is an apple)

5 apples- $1.30
1 apple- 1.30 / 5
           = $0.26
7 0
3 years ago
Name the figure using the correct symbol
Lelu [443]
A triangle a square and a rectangle
7 0
3 years ago
Given the following information about glucose levels (in milligrams of glucose per 100 milliliters of blood.) ( from 9.5)
Svetllana [295]

Answer:

A. 16.385

B. 9.532

C. (16.358)^2 is different from (9.532)^2

a. sigma squared of non pregnant women is GREATER THAN sigma squared of pregnant women

Step-by-step explanation:

<h3>A and B. Standard Deviation 's' of both columns:</h3>

The formula for the standard deviation is:

s = √(∑(x - μ)²/(n))

where,

∑ = is the sum function

x = a number from the set

μ  = mean of the set

n =  is the amount of the numbers in the set.

the column of non-pregnant women is

[73, 61, 104, 75, 85, 65, 62, 98, 92, 106]

  • first find the mean of this set:

mean = μ  = (sum of all the numbers)/(amount of numbers)

μ = [73 + 61 + 104 +75 + 85 + 65 + 62 + 98 + 92 + 106]/(10)

μ = 821/10

μ = 82.1

similarly, for pregnant women the mean is

μ = 80.125

  • now, to find the standard deviation 's', first we need to find the variance 's²'. So, what you have to do is subtract each value in the column with the mean 'μ' and square the result.

for non-pregnant you will get:

[ 82.81, 445.21, 479.61,  50.41,   8.41, 292.41, 404.01, 252.81,  98.01, 571.21]

for pregnant you will get:

[ 66.015,  15.015,  97.515, 221.265, 199.51,  102.515,   1.265,  23.76]

  • finally just sum all the numbers of a column and divide with the amount of numbers in that column (remember that col1 has 10 numbers and col2 has 8 numbers). And you will get your variance for both pregnant and non-pregnant women:

for non-pregnant = 2684.89/10

non-pregnant = 268.4 (this is the variance and it is denoted by s²)

for standard deviation 's', just take the square root of the variance

\sqrt{268.4} = 16.385.

similarly standard deviation of pregnant women can be found to be:

\sqrt{90.859} = 9.532.

A. non-pregnant 'S' =  16.385

B. pregnant 'S' = 9.532

<h3>C. CLAIM:</h3>

you only have to show whether the variance of the above two columns are different or not.

And YES, the variances of the two column are indeed different, hence you make the CLAIM as written in question(C)

<h3>Multiple Choice:</h3>

here you need to show how different are the two values:

recall the variances:

Column1 = 268.4 (non-pregnant)

Column2 = 9.532 (pregnant)

now you know that the variance of non-pregnant is GREATER THAN the variance of pregnant

3 0
3 years ago
Complete the function table for the function: y=4x+4<br> х<br> у<br> -6<br> -1<br> 2<br> 8
Shalnov [3]

Answer:

-6,-1

Step-by-step explanation:

y=4x+4 then simplify and came up with -6,-1.

3 0
4 years ago
What's 36y - 8y + 15 - 13?
Len [333]
The answer is 28y+2. You have to combine a like terms.
8 0
3 years ago
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