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2 years ago

What is 591.3 divided by 29?

Mathematics

2 answers:

vaieri [72.5K]2 years ago

6 0

The answer is 20.3896551724

DerKrebs [107]2 years ago

6 0

20.3896552, which rounds to 20.39, or 20.4
or just 20 if you're looking for whole numbers

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Subtract (8x + 9) - (6x - 1)

il63 [147K]

Answer:

=2x+10

Step-by-step explanation:

Let's simplify step-by-step.

8x+9−(6x−1)

Distribute the Negative Sign:

=8x+9+−1(6x−1)

=8x+9+−1(6x)+(−1)(−1)

=8x+9+−6x+1

Combine Like Terms:

=8x+9+−6x+1

=(8x+−6x)+(9+1)

=2x+10

Answer:

=2x+10

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3 years ago

(5 + w)(w + 4) = Need help not sure how to do it

Svetllana [295]

Answer:

w^2+ 9 + 2 0

Step-by-step explanation:

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3 years ago

Need help with all them, state how you know also Triangle Congruency

nata0808 [166]

Answer

where is the pdf?

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

It took Marjorie 15 minutes to drive from her house to her daughter's school. If they school was 4 miles away from her house, wh

notka56 [123]

The answer would be 3.75 miles por minute

8 0

2 years ago

Read 2 more answers

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