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3 years ago

Is 4/7 bigger or smaller than 3/5

Mathematics

2 answers:

crimeas [40]3 years ago

4 0

4/7=.57

3/5=.6

.6 is bigger than .57 so 4/7 is smaller than 3/5

blondinia [14]3 years ago

3 0

$\frac{4}{7}=\frac{4 \times 5}{7 \times 5}=\frac{20}{35} \\
\frac{3}{5}=\frac{3 \times 7}{5 \times 7}=\frac{21}{35} \\ \\
\frac{20}{35} < \frac{21}{35} \\ 
\Downarrow \\
\frac{4}{7} < \frac{3}{5}$

4/7 is smaller than 3/5.

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2 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

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(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

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

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

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3 years ago