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3 years ago

Equivalent expression of (m^1/3m^1/5)^0

Mathematics

1 answer:

Lapatulllka [165]3 years ago

7 0

Answer:

$\large\boxed{\left(m^\frac{1}{3}m^\frac{1}{5}\right)^0=1\ \text{if}\ m\neq0}$

Step-by-step explanation:

$\text{We know}\\\\a^0=1\ \text{for all value of}\ a\ \text{except 0}\\\\\left(m^\frac{1}{3}m^\frac{1}{5}\right)^0=1\ \text{if}\ m\neq0$

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MariettaO [177]

Answer:

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Step-by-step explanation:

Given: uniform distribution with min and max values of 18 and 33 respectively.

To find : probability density for upper cumulative frequency i.e 25.8 at least means x $\geq$ 25.8 upto 33 cm i.e the maximum limit of function.

Solution:

we have by definition , of uniform distribution

we get , probability density function defines as :

<em>F(x,a,b)= </em> $\frac{1}{b-a}$ <em> </em> $a\leq x\leq b$

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this is probability density function.

here the x=25.8 , a=18 and b=33

for lower cumulative frequency it defines as ;

P(x,a,b)= $\frac{x-a}{b-a}$ =25.8-18/33-18=0.52

for upper cumulative frequency it defines as ;

Q(x,a,b)=b-x/b-a=33-25.8/33-18=0.48

here at least 25.8 cm probability means it should be greater than a value(18cm) hence it is provided by the upper cumulative frequency

i.e. Q(x,a,b)=0.48

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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<h2>Given :</h2>

height (h) = 2 yd

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Answer:

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