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3 years ago

What is the value of x?

Mathematics

1 answer:

qaws [65]3 years ago

3 0

Answer:

Step-by-step explanation:

6x+12 = 9x-3

3x = 15

x= 5

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Please help this is due ASAP

Alik [6]

Answer:

Pretty sure its the third option

Step-by-step explanation:

8 0

3 years ago

Wat is $3.459B 2.443B $55.838B 85.965B in dollars pls

Oksanka [162]

Answers:

$3,459,000,000

$2,443,000,000

$55,838,000,000

$85,965,000,000

8 0

3 years ago

mr cold war is not here... when it comes to Fractions: what the f... is 6/8×2/4×7/15=?

ryzh [129]

Your answer is 7/40 or 0.175

8 0

3 years ago

sorry again for asking this but can someone explain this and give me the answer. Been having a hard time learning this.

Rudik [331]

Hey!!!

In question no.4, we have to subtract the equation.

In question no.5 we have to add the equation.

Hope it will help you...

<h3>Have a great day</h3>

3 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