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olga nikolaevna [1]
3 years ago
14

When would there be only four different equations for a set of math mountain numbers

Mathematics
2 answers:
Yuri [45]3 years ago
7 0
It is called a fact family

RSB [31]3 years ago
4 0
The name it is called a fact family
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Find an equation of the line passing through the points (5, 21) and (-5, -29)
gogolik [260]
Trata 5-29+-5=21 porque ya as tenido este
4 0
3 years ago
PLEASE HELP!!! Find area of each square
Illusion [34]

Answer:

25 for the top right, 16 for the left, 9 for the bottom

Step-by-step explanation:

area = length times width

6 0
3 years ago
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Simplify 4(x+2y)-(3x-8y)​
stealth61 [152]

Answer: x+16y

Step-by-step explanation:

4(x+2y)-(3x-8y)​

4x+8y-3x-8y

x+16y

7 0
3 years ago
A sample of 173 students using Method 1 produces a testing average of 88. A sample of 127 students using Method 2 produces a tes
Lerok [7]

Answer:

43 < μ₁ - μ₂  < 49

Step-by-step explanation:

Sample 1:

Mean      x₁    =  88

Sample size    n₁  = 173

Sample standard deviation    s₁  =  12,22

Sample 2:

Mean      x₂    =  53,5

Sample size    n₂  = 127

Sample standard deviation    s₂  =  13,59

CI   95 %        α  =  5%    α = 0,05    α/2  = 0,025

We need to find:

(  x₁  -  x₂  ) - tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂) < μ₁ - μ₂  < (  x₁  -  x₂  ) + tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂)

v = degree of fredom    

v  = [ ( s₁²/n₁ + s₂²/n₂)² / (s₁²/n₁)² /n₁-1  + (s₂²/n₂)²/n₂-1

v =  [ (0,86 + 1,45 ) / 0,0043 +  0,017

v = 2,31 / 0,0213

v = 108

Then  t 0,025, 108  from t table is:  We will take v = 100

t = 1,984

Now

√ (s₁²/n₁ ) + (s₂²/n₂)  =  √ 0,86  + 1,43

√ 2,3   = 1,51

Then: CI:

(173 - 127 ) -  1,984*1,51    < μ₁ - μ₂  < ( 173 - 127 )  +  1,984*1,51

46 - 3  < μ₁ - μ₂  < 46 + 3

43 < μ₁ - μ₂  < 49

8 0
3 years ago
solve the system of equations y = 4x + 1 y = x^2 + 2x - 2 A. (-3,-13) and (1,3) B. (-3,13) and (1,-3) c. (1,3) and (-3,13) D. (-
sergejj [24]

Answer:

option D

(-1,-3) and (3,13)

Step-by-step explanation:

Given in the question two equations

y = 4x + 1

y = x² + 2x - 2

Equate both functions

4x + 1 = x² + 2x - 2

rearrange the x term and constant

-x² + 4x - 2x + 2 + 1 = 0

-x² + 2x + 3 = 0

factors

-x * 3x = -3x²

-x + 3x = 2x

-x² -x + 3x + 3 = 0

-x(x+1) +3(x+1) = 0

solve

(x+1)(3-x) = 0

x = -1

and

x = 3

Plug these values in equation to find y

x = -1

y = 4(-1)+ 1

y = -3

x = 3

y = 4(3)+ 1

y = 13

7 0
4 years ago
Read 2 more answers
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