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2 years ago

5

What are the x and y-intercepts of the line described by the equation? -6x + 3y = 18.9

1 answer:

-BARSIC- [3]2 years ago

8 0

Answer:

x-intercept - x = 1/2y - 3.15

y-intercept - y = 2x + 6.3

Step-by-step explanation:

For x,

-6x + 3y = 18.9

-6x = -3y + 18.9

x = -3y/-6 + 18.9/-6

x = 1/2y - 3.15

For y,

-6x + 3y = 18.9

3y = 6x + 18.9

y = 6x/3 + 18.9/3

y = 2x + 6.3

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2 years ago

What are the coordinates of each point after quadrilateral ABCD is reflected across the y-axis and then translated 2 units down?

zheka24 [161]

Answer:

$A" =(-1,-1)$ $B" =(-2,1)$

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Step-by-step explanation:

Given

$A = (1,1)$ $B =(2,3)$

$C = (4,3)$ $D = (5,1)$

Required

The new coordinates after reflected across y-axis and shifted 2 units down

When a point is reflected across the y-axis, the rule is:

$(x,y)=>(-x,y)$

So:

$A = (1,1)$ $=> A' = (-1,1)$

$B =(2,3)$ $=> B' = (-2,3)$

$C = (4,3)$ $=> C' = (-4,3)$

$D = (5,1)$ $=> D' = (-5,1)$

When a point is translated 2 units down, the rule is:

$(x,y) => (x,y-2)$

So:

$A = (1,1)$ $=> A' = (-1,1)$ $=> A" =(-1,-1)$

$B =(2,3)$ $=> B' = (-2,3)$ $=> B" =(-2,1)$

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2 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

i)

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

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2 years ago

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