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Tju [1.3M]
2 years ago
15

Write an equation of the line that passes through (18,2) and is parallel to the line 3y - x= - 12 .

Mathematics
1 answer:
Gekata [30.6K]2 years ago
8 0

Answer:

y = \frac{1}{3}x-4

Step-by-step explanation:

<u>Step 1:  Solve for y in the first equation</u>

3y - x = -12

3y - x + x = -12 + x

\frac{3y}{3} = \frac{x}{3} - \frac{12}{3}

y = \frac{1}{3}x - 4

<u>Step 2:  Determine the important aspects</u>

We know that our line is parallel to the other line that has a slope of 1/3 which means that our slope is also going to be 1/3.  We also know that our line crosses the point (18, 2) which means that we can use the point slope form to determine our equation

Point Slope Form → (y-y_1) = m(x - x_1)

<u>Step 3:  Plug in the information and solve</u>

(y-2) = \frac{1}{3}(x - 18)

y - 2 = \frac{1}{3}x - 6

y - 2 + 2 = \frac{1}{3}x - 6 + 2

y = \frac{1}{3}x-4

Answer: y = \frac{1}{3}x-4

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Answer:

\therefore 2\dfrac{1}{3}  \div \left ( -3\dfrac{2}{3} \right ) = -\dfrac{7}{11}

Step-by-step explanation:

The question relates to dividing a fraction by proper and another fraction

The given expression is presented as follows;

2\dfrac{1}{3}  \div \left ( -3\dfrac{2}{3} \right )

We rearrange the fractions as improper  fractions for easier division as follows;

2\dfrac{1}{3}  = \dfrac{7}{3}

-3\dfrac{2}{3} = -\dfrac{11}{3}

Therefore, we have;

2\dfrac{1}{3}  \div \left ( -3\dfrac{2}{3} \right ) = \dfrac{7}{3} \div \left (-\dfrac{11}{3} \right ) = \dfrac{7}{3} \times \left (\dfrac{1}{-\dfrac{11}{3} } \right ) = \dfrac{7}{3} \div \left (-\dfrac{3}{11} \right ) =-\dfrac{7}{11}

\therefore 2\dfrac{1}{3}  \div \left ( -3\dfrac{2}{3} \right ) = -\dfrac{7}{11}

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3 years ago
Flat screen tv cost $1,400 the tv is on sale 35% off what is the discounted price on the tv
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Easy,

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Multiply the list price ($1,400) by the complement.

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Thus, the new price is, $910.
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. ​(a) Describe the
Ivanshal [37]

Answer:

a) \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

b) z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

c) z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

d) z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

\mu = 83, \sigma = 27

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})

With:

\mu_{\bar X}= 83

\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3

Part b

We want this probability:

P(\bar X>89)

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 89 we got:

z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2

P(Z>2) = 1-P(Z

Part c

P(\bar X

We can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score for 75.65 we got:

z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45

P(Z

Part d

We want this probability:

P(79.4 < \bar X < 89.3)

We find the z scores:

z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1

z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2

P(-1.2

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4 years ago
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