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3 years ago

Is (2, 2) a solution of y < 4x - 6? Choose 1 answer yes or no

Mathematics

1 answer:

zalisa [80]3 years ago

4 0

No, (2,2) is not solution of y < 4x - 6

Step-by-step explanation:

To solve above linear problem, we use standard method of Back- Substitution.

First of all, A point (2,2) only becomes solution of equation, if it satisfies equation $y < 4x - 6$ .

Here, the $y < 4x - 6$ seems to be like a linear equation.

In this Equation, using Back- Substitution method,

Point (2,2) : first digit 2 corresponds to x- coordinate and second digit 2 corresponds to y-coordinate.

$y < 4x - 6.....(1) \\\ x= 2 .....(2) \\\ y= 2 .....(3)$

By putting value of x and y in equation (1),

Equation becomes,

$2 < 4 \times 2 - 6$

$2 < 8 - 6$

$2 < 2$ .... not true mathematically.

Finally, the result came is not true. Therefore, (2,2) is not solution for equation

$y < 4x - 6$

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19. Find the value of X (In the picture) (giving points to best answer/brainlest)

Tema [17]

Answer:

x = 61 degrees

Step-by-step explanation:

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Also the left triangle is isosceles, so we have

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3 years ago

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Hatshy [7]

Answer:

The bait is 14 feet below the water surface

Step-by-step explanation:

Here, we want to know the final position of the bait relative to the surface of the water.

Firstly, he starts with the bait at 2 feet below the water surface. The reels out the bait a further 19 feet, the new position of the bait relative to the water surface will be 2 + 19 = 21 feet

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

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