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3 years ago

Write the equation in slope-intercept form: 3y = 2x +15

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

Answer:

y = 2/3x + 5

Step-by-step explanation:

Hope this helps! :)

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Question: "Solve for volume and round to the nearest tenth, if necessary use 3.14."

ira [324]

When I solved it I got 26648.8

3 0

3 years ago

The ordered pair (3,9) is a solution to the system of

nadya68 [22]

Answer:

(2,7) is not a solution to the given system of equations.

Step-by-step explanation:

Given system of equation is:

2x + 3 = y

2x + y = 15

To check whether (2,7) is solution to this system or not, we will put x=2 and y=7 in both equations.

Putting x=2 and y=7 in Eqn 1

2(2) + 3 = 7

4 + 3 = 7

7 = 7

Thus the ordered pair satisfies the equation

Putting x=2 and y=7 in Eqn 2

2(2) + 7 = 15

4 + 7 = 15

11 ≠ 15

The ordered pair do not satisfy the second equation.

Hence,

(2,7) is not a solution to the given system of equations.

8 0

3 years ago

What is the area of the polygon defined by the points A(12, 20), B(20, 20), C(20, 8), and D(12, 8)? a 96 square units b 132 squa

Leto [7]

Answer:

a. 96 square units

Step-by-step explanation:

The figure is a rectangle with width AB = (20-12) = 8 units and height BC = (20-8) = 12 units.

The area of the rectangle is (8 units)×(12 units) = 96 square units.

3 0

3 years ago

M=F+Frt<br><br> solve for t...

Aliun [14]

Answer:

m=f+Frt

M=f(1+rt)

M\1+rt=F

4 0

3 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

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

6 0

3 years ago