How do i solve this step by step? 3( q

X=1/16y^2 the directrix of the parabola is

yan [13]

To solve this problem you must apply the proccedure shown below:

1. You have the following equation of a parabola, given in the problem above:

x=1/16y^2

2. Then, based on the graph attached, you have:

p=y^2/4x
p=8^2/(4)(4)
p=64/16
p=4

3. The directrix is:

directrix=h-p
directrix=0-4
directrix=-4

The answer is:-4

3 0

3 years ago

Do the data in the table represent a direct variation or inverse variation? Write an equation to model the data in the table? x

pantera1 [17]

Answer:

hauawiwiwiwiwiwiwiwiwiwiwuwjhehehwhwe

7 0

3 years ago

Read 2 more answers

Factor −5x2 + 10x. −5x(x + 2) 5(−x2 + 10x) 5x(−x + 2) x(5x + 10)

Hitman42 [59]

${ \red{ \bold{option \: (c)}}}$

Step-by-step explanation:

${ \purple{ \tt{ { - 5x}^{2} + 10x}}}$

Take ${ \purple{ \tt{5x}}}$ as common. Then,

${ \purple{ \tt{5x( - x + 2)}}}$

4 0

1 year ago

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138