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3 years ago

9

Solve the linear system by graphing y = x - 1 y =3x - 9

1 answer:

Zielflug [23.3K]3 years ago

4 0

I think is
y = x - 1y = 3x - 9
y - y = x - 3x = -9
0 = - 2x = - 9

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Explain how to use a horizontal number line<br> to order four fractions from least to greatest.

horsena [70]

Answer:

To order fractions from least to greatest, start by finding the lowest common denominator for all of the fractions. Next, convert each of the fractions by dividing the lowest common denominator by the denominator and then multiplying the top and bottom of the fraction by your answer.

Step-by-step explanation:

Solution for the above problem: We first find the least common denominator by finding the least common multiple for 4, 3, 2, 6, and 8. We find the LCM by the prime factorization method. Next, place the fractions in order from least to greatest.

4 0

3 years ago

Given: F(x) = 5x - 6 and G(x) = x - 4

fomenos

Answer:

Correct choice is $\dfrac{x+26}{5}$

Step-by-step explanation:

1. If $F(x)=5x-6,$ then you can find the inverse $F^{-1}(x):$

$y=5x-6,\\ \\5x=y+6,\\ \\x=\dfrac{y+6}{5},\\ \\F^{-1}(x)=\dfrac{x+6}{5}.$

2. If $G(x)=x-4,$ then

$y=x-4,\\ \\x=y+4,\\ \\G^{-1}(x)=x+4.$

3. Hence,

$G^{-1}(F^{-1}(x))=G^{-1}\left(\dfrac{x+6}{5}\right)=\dfrac{x+6}{5}+4=\dfrac{x+26}{5}.$

7 0

3 years ago

What is a possible value for the missing term of the geometric sequence ?

lesantik [10]

50 * 3 = 150 and 150 * 3 = 450

So the answer is b 150

3 0

3 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

Please help 10 points

Natasha_Volkova [10]

The following applies:

90°

180°

270°

3 0

3 years ago