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2 years ago

04 Name the

Mathematics

1 answer:

kirill [66]2 years ago

5 0

Answer:

<h3>integers integers OK ok ok ok ok ok ok</h3>

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How do you solve -2+6(-6x+6)?

Natali5045456 [20]

Answer: −36x+34

Step-by-step explanation:

Equation: −2+6(−6x+6)

Step 1. distribute the equation:

−2+(6)(−6x)+(6)(6)

−2+−36x+36

Step 2. Finally combine like terms:

−2+−36x+36

(−36x)+(−2+36)

−36x+34

Final Answer:

−36x+34

Hope I could help! :)

4 0

3 years ago

Help me or give me answer key plz I’m doing Desmos Quadratic Transformations – Part 1 I’m dum

Nina [5.8K]

Answer:

While the slider moves to the right the point change also. You can see that the points will decrease and increase again.

7 0

2 years ago

Help please i need it thanks.

Shalnov [3]

Answer:

x is less than or equal to 7

Step-by-step explanation:

8 0

2 years ago

What is the slope of y-9=3(x-2) A. 3 B. 1/2 C. 2 D. 1/9

pav-90 [236]

Answer:

A. 3

Step-by-step explanation:

y - 9 = 3(x-2)

y = 3x - 6 + 9

y = 3x + 3

slope is 3x = 3

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