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help please!
Prove that,
when the values in a database are equal to each other, then the A.M, G.M and H.M equal to each other
note:
A.M=arithmetic mean
G.M=geometric mean
H.M= harmonic mean​


Sagot :

Answer:

See below

Step-by-step explanation:

the n number of value of x

[tex] \displaystyle x_{1},x _{2} \dots x_{n}[/tex]

let it be

[tex] \displaystyle x_{1} = x _{2} = x_{3}{\dots }= x_{n} = a[/tex]

now, the A.M of x is

[tex] \rm \displaystyle \: A.M = \frac{ x_{1} + x_{2} + \dots \dots \: + x_{n} }{n} [/tex]

since every value equal to a

substitute:

[tex] \rm \displaystyle \: A.M = \frac{ a + a + \dots \dots \: + a}{n} [/tex]

[tex] \rm \displaystyle \: A.M = \frac{ na}{n} [/tex]

reduce fraction:

[tex] \rm \displaystyle \: A.M = a[/tex]

the G.M of x is

[tex] \rm\displaystyle \: G.M =( x_{1} \times x _{2} {\dots }\times x_{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

since every value equal to a

substitute:

[tex] \rm\displaystyle \: G.M =( a \times a{\dots }\times a{)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M =( {a}^{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M = a[/tex]

the H.M of x is

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ x_{1}} + \frac{1}{ x_{2} } {\dots } \: { \dots}\frac{1}{x _{n} } } [/tex]

since every value equal to a

substitute:

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ a} + \frac{1}{ a } {\dots } \: { \dots}\frac{1}{a } } [/tex]

[tex] \displaystyle \: H.M = \frac{n}{ \dfrac{n}{a} } [/tex]

simplify complex fraction:

[tex] \displaystyle \: H.M = n \times \frac{a}{n} [/tex]

[tex] \displaystyle \: H.M = a \: [/tex]

so

[tex] \displaystyle \: A.M = G.M = H.M = a[/tex]

hence,

[tex]\text{Proven}[/tex]

Answer:

What [tex]\colorbox{red}{Nayefx}[/tex]says is I say