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3 years ago

Find the slope of (3,-3) and (0,3)

Mathematics

1 answer:

Vinil7 [7]3 years ago

7 0

Answer:

<h3> $\boxed{ \bold{ \huge{ \boxed{ \sf{ - 2}}}}}$ </h3>

Step-by-step explanation:

Let the points be A and B

A ( 3 , -3 ) ⇒( x₁ , y₁ )

B ( 0 , 3 )⇒ ( x₂ , y₂ )

<u>Finding</u><u> </u><u>the </u><u>slope</u>

Slope ( m ) = $\sf{ \frac{y2 - y1}{x2 - x1} }$

plug the values

⇒ $\sf{ \frac{3 - ( - 3)}{0 - 3} }$

We know that , $\sf{( - ) \times ( - ) = ( + )}$

⇒ $\sf{ \frac{3 + 3}{0 - 3} }$

Calculate

⇒ $\sf{ \frac{6}{ - 3} }$

⇒ $\sf{ - 2}$

Hope I helped!

Best regards!!

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3 years ago

Savatey [412]

Answer:

Step-by-step explanation:

(X1, Y1) = (-3,-2)

(X2,Y2)=(4,8)

SLOPE= (Y2-Y1) / (X2-X1)

= (8-(-2)) / (4-(-3))

=(8+2) / (4+3)

= 10/7

Hope it was helpful

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3 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

3 years ago