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3 years ago

Which choice could be the equation of a line parallel to the line represented by this equation 3x-2y=6 ?

1 answer:

6 0

You would change what the problem is equal to, so it could be 3x-2y=4 or any other answer as long as the coefficients are not changed.

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The graph of a line has a slope of -3 and passes through the point (-6,-4).

prohojiy [21]

Answer:

(A) y+4=-3(x+6)

Step-by-step explanation:

The point-slope form of the equation of a line whose slope is m and passes through the point $(x_1, y_1)$ is: $y-y_1=m(x-x_1)$

Given the point: $(x_1, y_1)=(-6,-4)$

Slope, m=-3

$x_1=-6\\y_1=-4$

Substituting these values into: $y-y_1=m(x-x_1)$ , we obtain the point slope form of the equation:

$y-(-4)=-3(x-(-6))\\\\y+4=-3(x+6)$

The correct option is A.

6 0

3 years ago

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

Solve by taking the square root of both sides 3(x-1)^2-162=0

Natali5045456 [20]

We know that: 3(x-1)^2-162=0
or (x-1)²= 162:3
and (x-1)²= 54
we take the square root of both sides
* x-1=√54= 3√6 or x=1+3√6
* x-1= -√54= -3√6 or x=1-3√6
This equation has 2 solutions

7 0

3 years ago

Read 2 more answers

The product of two consecutive, negative integers is 182. What equation represents the situation?

Sholpan [36]

-13 and -14

They both are consecutive negative integers that multiply to 182.

4 0

3 years ago

Read 2 more answers

Write ANY FOUR factors of -23 ab2.

Sauron [17]

Answer:

The factors are;

23, -23 , a , b

Step-by-step explanation:

Here, we want to write the factors of the given expression

-23 ab^2

The factors are simply those values or expression that can be divided by the given expressions

We have the factors as follows;

-23

7 0

3 years ago