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3 years ago

Which graph represents the inequality y > 3 − x ?

Mathematics

1 answer:

viktelen [127]3 years ago

6 0

To solve this problem you must think in this way.

First we see that function y is greater than 3-x, conclusion is the following

shaded area must be above function. That condition is satisfied by the graph 3 and 4.

Then we write inequality in the slope form y> -x+3.

We see that the coefficient of direction is -1 (negative) which mean that the graph is decreasing. That condition is satisfied by the graph No 4.

The correct answer is graph number 4 ( or D.)

Good luck!!!

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What is the result if you divide 18r4 s5 t6 / -3 r2 st3 ?

coldgirl [10]

The answer for the given above is letter "a, -6r2s4t3". This is obtained by dividing 18 by -3 and subtracting the exponent below with that from above of the same variable. For r, subtract 2 from 4 to get 2. For s, subtract 1 from 5 to get 4 and lastly for t, subtract 3 from 6 to get 3.

5 0

3 years ago

Read 2 more answers

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:

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