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3 years ago

What is equivalent to 4/3 that is not an improper fraction

Mathematics

1 answer:

telo118 [61]3 years ago

4 0

Answer:

1 1/3

Step-by-step explanation:

Something that is not a inproper fraction is a proper fraction which is :

The numerator is less than the denominator

Examples: 1/3, 3/4, 2/7.

Or a mixed fraction which is:

A whole number and proper fraction together

Examples: 1 1/3, 2 1/4, 16 2/5

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Answer:

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Step-by-step explanation:

Given

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$h = 2$ --- height

Required

The volume (V) of the cylinder

This is calculated using:

$V = \pi r^2h$

So, we have:

$V = \frac{22}{7} * 2^2 * 2$

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Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

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