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kkurt [141]
3 years ago
14

How to make an equation out of perimeter and area

Mathematics
2 answers:
Natasha2012 [34]3 years ago
7 0
Where P = perimeter, L = length, W = width, and A = area:

A = LW
P = 2(LW)

Hope this helped! :)
QveST [7]3 years ago
6 0
Area=Length*Width
Perimeter=2(Length+Width)
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A system of linear equations with fewer equations than unknowns is sometimes called an underdetermined system. Can such a system
elena55 [62]

Answer:

No, it cannot have a unique solution. Because there are more variables than​ equations, there must be at least one free variable. If the linear system is consistent and there is at least one free​ variable, the solution set contains infinitely many solutions. If the linear system is​ inconsistent, there is no solution.

Step-by-step explanation:

the questionnaire options are incomplete, however the given option is correct

We mark this option as correct because in a linear system of equations there can be more than one solution, since the components of the equations, that is, the variables are multiple, leaving free variables which generates more alternative solutions, however when there is no consistency there will be no solution

3 0
3 years ago
What is 312÷2 can someone help
Kisachek [45]

156

Divide using a calculator or long division

3 0
2 years ago
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−x=5<br> x =<br><br><br><br> wut is thi$
Vlada [557]

Answer:

x = -5

Step-by-step explanation:

-x = 5

*-1   *-1

-x * -1 = x

5 * -1 = -5

x = -5

5 0
3 years ago
Daniel deposits $300 into an account that earns 16% interest annually. Which equation can be used to model his account balance,
MAVERICK [17]

Answer:

y=300(1+0.16)^x

Step-by-step explanation:

<h2>This account  can be  modeled using the compound interest formula.</h2><h2>the compound interest formula is expressed as</h2>

 A= P(1+r )^t

Where

A =final amount  = y

P=initial principal balance = $300

r=interest rate  = 16%= 0.16

t=number of time periods elapsed= x

Hence the equation to model his account balance/ final amount A (y) after time (x) years is

y=300(1+0.16)^x

5 0
2 years ago
onsider the following hypothesis test: H 0: 50 H a: &gt; 50 A sample of 50 is used and the population standard deviation is 6. U
kondaur [170]

Answer:

a) z(e)  >  z(c)   2.94 > 1.64  we are in the rejection zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12  > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

                                          b) Hₐ    μ > 50   ;     for value 51

                                          c) Hₐ    μ > 50   ;      for value 51.8

With value 52.5

The test statistic    z(e)  ??

a)  z(e) =  ( μ  -  μ₀ ) /( σ/√50)      z(e) = (2.5*√50 )/6   z(e) = 2.94

2.94 > 1.64  we are in the rejected zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) =  ( μ  -  μ₀ ) /( σ/√50)    ⇒  z(e) =  √50/6    ⇒  z(e) = 1.18

z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

8 0
3 years ago
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