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mars1129 [50]
3 years ago
13

Which equation represents the line that passes through the points (-5,2) and (9,0)

Mathematics
1 answer:
V125BC [204]3 years ago
7 0

Step-by-step explanation:

<em>Given </em><em>points </em>

<em>(</em><em> </em><em>-</em><em>5</em><em> </em><em>,</em><em> </em><em>2</em><em>)</em><em> </em><em>and </em><em>(</em><em>9</em><em> </em><em>,</em><em> </em><em>0</em><em>)</em>

<em>First </em><em>calculating </em><em>slope(</em><em>m</em><em>)</em>

<em>=</em><em> </em><em>(</em><em> </em><em>y2 </em><em>-</em><em> </em><em>y1</em><em>) </em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>x2 </em><em>-</em><em> </em><em>x1</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>0</em><em>-</em><em>2</em><em>)</em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>9</em><em>+</em><em>5</em><em>)</em>

<em>=</em><em> </em><em>-</em><em>2</em><em> </em><em>/</em><em> </em><em>1</em><em>4</em>

<em>=</em><em> </em><em>-</em><em>1</em><em>/</em><em>7</em>

<em>Now </em><em>The </em><em>equation </em><em>of </em><em>line </em><em>is </em><em>given </em><em>by </em>

<em>y </em><em>-</em><em> </em><em>y1 </em><em>=</em><em> </em><em>m </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>x1) </em>

<em>y </em><em>-</em><em>2</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em><em>/</em><em>7</em><em> </em><em>(</em><em>x </em><em>+</em><em> </em><em>5</em><em>)</em>

<em>7</em><em>(</em><em> </em><em>y </em><em>-</em><em> </em><em>2</em><em>)</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>5</em><em>)</em>

<em>7y </em><em>-</em><em> </em><em>1</em><em>4</em><em> </em><em>=</em><em> </em><em>-</em><em>x </em><em>-</em><em>5</em>

<em>x </em><em>+</em><em> </em><em>7y </em><em>-</em><em>1</em><em>4</em><em> </em><em>+</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em>

<em>x </em><em>+</em><em> </em><em>7y </em><em>-</em><em> </em><em>9</em><em> </em><em>=</em><em> </em><em>0</em><em> </em>

<em>Which </em><em>is </em><em>the </em><em>required </em><em>equation </em>

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13x + 15=17x-13 solve for the solution
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Answer: x=7

Step-by-step explanation:

Step 1: Subtract 17x from both sides.

13x+15−17x=17x−13−17x

−4x+15=−13

Step 2: Subtract 15 from both sides.

−4x+15−15=−13−15

−4x=−28

Step 3: Divide both sides by -4.

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Answer:

m∠b = 35°

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Step-by-step explanation:

To find m∠b, we use Definition of Supplementary Angles:

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Step-by-step explanation:

\sqrt{x}  \cdot \sqrt{x}  = \sqrt{x^2}  = x^{ 2 \times \frac{1}{2}} = x

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Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential dis
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a) P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

b) P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}, x>0

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

X \sim Exp(\lambda=\frac{1}{30000})

The cumulative distribution function is given by:

F(X) = 1- e^{-\frac{x}{\mu}}

Part a

We want to find this probability:

P(X>30000) and for this case we can use the cumulative distribution function to find it like this:

P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368

Part b

For this case w want to find this probability

P(X>30000|X>15000)

We have an important property on the exponential distribution called "Memoryless" property and says this:

P(X>a+t| X>t)=P(X>a)  

On this case if we use this property we have this:P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)

We can use the definition of the density function and find this probability:

P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607

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