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Arlecino [84]
4 years ago
9

Inverse variation?

Mathematics
1 answer:
Alex_Xolod [135]4 years ago
3 0

Answer:

An Inverse variation states that a relationship between two variables in which the product is always constant.

If one of the variable increases then the other decreases in proportion so that the product is unchanged.

if y is inversely proportional to x, then the equation is of the form

y = \frac{k}{x} , where k is constant.

As per the given condition, A constant force acts upon an object, the acceleration of the object varies inversely with it's mass.

i.e, a \propto \frac{1}{m}

By definition of inverse variation;

a = \frac{F}{m} . ......[1]  ,where F is the constant force.

It is given that an object has mass(m)= 4kg and the acceleration of an object(a) = 14 m/s^2

Substitute these in [1] to find the constant force F;

14 = \frac{F}{4}

or

F =14 \times 4 = 56 kg m/s^2

Now, if the same force acts upon the another object whose mass (m) = 7 kg then calculate acceleration.

[1]⇒  a = \frac{56}{7} = 8 m/s^2

therefore, the acceleration of an another object is, 8 m/s^2


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umka2103 [35]

Answer:

a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

X is a random variable of Poisson distribution with mean \mu = 3 , and Y is a random variable with range {0, 1, 2, 3, 4}, with the following values

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we obtain:

PY(0) = ε^(-3) = 0.04978..

PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

So, if we ask for 6 newspapers, the probability of receiving at least 6 calls is 0.966489, and the probability to receive more calls than available newspapers will be less than 0.05.

I hope this helped you!

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