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2 years ago

What is 15 meters decreased by 60%

Mathematics

2 answers:

expeople1 [14]2 years ago

5 0

Answer:

-9

Step-by-step explanation:

MArishka [77]2 years ago

3 0

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{60\% of 15}}{\left( \cfrac{60}{100} \right)15}\implies 9~\hspace{7em} \stackrel{15-9}{6}$

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You have 42 ft. of fencing (1 ft. segments) to make a rectangular garden. How much should each side be to maximize your total ar

Verizon [17]

Answer:

$Length=10.5\ ft$

$Width=10.5\ ft$

$Area=110.25\ ft^{2}$

Step-by-step explanation:

Let

x----> the length of the rectangular garden

y---> the width of the rectangular garden

we know that

The perimeter of the rectangle is equal to

$P=2(x+y)$

we have

$P=42\ ft$

$42=2(x+y)$

simplify

$21=(x+y)$

$y=21-x$ ------> equation A

Remember that the area of rectangle is equal to

$A=xy$ ----> equation B

substitute equation A in equation B

$A=x(21-x)$

$A=21x-x^{2}$ ----> this is a vertical parabola open downward

The vertex is a maximum

The y-coordinate of the vertex is the maximum area

The x-coordinate of the vertex is the length side of the rectangle that maximize the area

using a graphing tool

The vertex is the point $(10.5,110.25)$

see the attached figure

$x=10.5\ ft$

Find the value of y

$y=21-10.5=10.5\ ft$

The garden is a square

the area is equal to

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6 0

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]

Answer:

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

6 0

3 years ago