{y=3/4x 5/2x+2y=5 need help

1) A research team is planning to use an ice drill at the top of a glacier. The team plans to use the drill for 10 hours each da

MaRussiya [10]

Answer:

Answer is 50 trust me

Step-by-step explanation:

7 0

3 years ago

Ms. Reynold's sprinkler system has 5 stations that water all parts of her front and back lawn. Each station runs for an equal am

mina [271]

Answer: 40 minutes

Step-by-step explanation:

It takes 16 minutes for the first 2 stations to water their sections which means that the time taken per station is:

= 16/2

= 8 minutes per station

Ms. Reynolds has 5 stations.

Total time taken will therefore be:

= 8 * 5

= 40 minutes

3 0

3 years ago

PLEASE ANSWER AND HELP

ra1l [238]

She must save $91.80.

7 0

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

Use the functions to find x

suter [353]

Answer:

x = 14

Step-by-step explanation:

$f(x) = 10 - \frac{9}{2} x \\ \\ plug \: f(x) = - 53 \\ \\ - 53 = 10 - \frac{9}{2} x \\ \\ \frac{9}{2} x = 10 + 53 \\ \\ \frac{9}{2} x = 63 \\ \\ x = 63 \times \frac{2}{9} \\ \\ x = 7 \times 2 \\ \\ x = 14$

5 0

3 years ago