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3 years ago

1/2x + 3y=4 for x when x = 6

Mathematics

2 answers:

Kipish [7]3 years ago

6 0

Wouldn't it be y=1/3?

WARRIOR [948]3 years ago

4 0

½6+3y=4
3+3y=4
3y=4-3
3y=1
y=⅓

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Leto [7]

Answer:

52/99

Step-by-step explanation:

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3 years ago

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Which equations are equivalent to Y=2/3X-4 when written in slope-intercept form? Check all that apply. A =3X-2Y=4. B=2X-3Y=12. C

klemol [59]

Answer:

B and C

Step-by-step explanation:

Rearrange each equation into slope- intercept form

3x - 2y = 4 ( subtract 3x from both sides )

- 2y = - 3x + 4 ( divide all terms by - 2 )

y = $\frac{3}{2}$ x - 2 ← not equivalent

2x - 3y = 12 ( subtract 2x from both sides )

- 3y = - 2x + 12 ( divide all terms by - 2 )

y = $\frac{2}{3}$ x - 4 ← equivalent

- 4(2x - 3y ) = - 48 ( divide both sides by - 4 )

2x - 3y = 12 ← same as B ⇒ equivalent

2(x + 6) = 3y

3y = 2x + 12 ( divide all terms by 3 )

y = $\frac{2}{3}$ x + 4 ← not equivalent

2x - 3y = 4 ( subtract 2x from both sides )

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y = $\frac{2}{3}$ x - $\frac{4}{3}$ ← not equivalent

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3 years ago

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Assume that the Poisson distribution applies and that the mean number of aircraft accidents is 9 per month. Find P(0), the prob

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Answer: 0.0001

It is unlikely to have a month with no aircraft accidents .

Step-by-step explanation:

Given : Mean number of aircraft accidents = 9 per month

The Poisson distribution formula :-

$\dfrac{e^{-\lambda}\lambda^x}{x!}$ , where $\lambda$ is the mean of the distribution.

If X = the number of aircraft accidents per month, then the probability that in a month, there will be no aircraft accidents will be :-

$\dfrac{e^{-9}(9)^0}{0!}=0.000123409804087\approx0.0001$

Hence, the probability that in a month, there will be no aircraft accidents = 0.0001

Since this is less than 0.5 , therefor it is unlikely to have a month with no aircraft accidents .

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3 years ago

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3 years ago

Given: Angle 2 is 65 degrees. What is the angle measurement of Angle 4. Explain the angle relationship used and show your work.

lesya [120]

Answer:

$\angle 4 = 115^{\circ}$

$\angle 3 = 65^{\circ}$

$\angle 7 = 65^{\circ}$

Step-by-step explanation:

Given

$\angle 2 = 65^{\circ}$

See attachment

Solving (a): $\angle 4$

To solve for $\angle 4$ , we make use of:

$\angle 2 +\angle 4 = 180^{\circ}$

The relationship between both angles is that they are complementary angles

Make $\angle 4$ the subject

$\angle 4 = 180^{\circ} - \angle 2$

Substitute $65^{\circ}$ for $\angle 2$

$\angle 4 = 180^{\circ} - 65^{\circ}$

$\angle 4 = 115^{\circ}$

Solving (b): $\angle 3$

To solve for $\angle 3$ , we make use of:

$\angle 3 =\angle 2$

The relationship between both angles is that they are complementary angles

$\angle 3 = 65^{\circ}$

Solving (c): $\angle 7$

To solve for $\angle 7$ , we make use of:

$\angle 7 = \angle 3$

The relationship between both angles is that they are alternate exterior angles.

So:

$\angle 7 = 65^{\circ}$

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3 years ago