6. What is the sum of the solutions?

What is the next number? 3,6,17,42,87,158,...,...

Alla [95]

Answer:

CAN'T TELL . BECAUSE I HAVEN'T SEEN SUCH A QUESTION BEFORE

5 0

2 years ago

A hovercraft takes off from a platform it’s height (in meters), x seconds after takeoff, is modeled by

yulyashka [42]

9514 1404 393

Answer:

81 m

Step-by-step explanation:

Since x is seconds after takeoff, the value of x at takeoff is 0. You are being asked to find H(0).

H(0) = -3(0 -3)^2 +108 = -3(9) +108 = 81

The height of the hovercraft at the time of takeoff is 81 meters.

8 0

3 years ago

Smoothing a time series of observations A. is a form of statistical cheating. B. allows statisticians to use less data than woul

DedPeter [7]

Answer:

D. is used to reveal an underlying pattern in the data.

Step-by-step explanation:

Smoothing a time series is achieved when a computer uses some pre-programmed calculation methods to remove noise from large volumes of data. Smoothing helps a user detect patterns in a set of data, thus making it possible to make future predictions. For example, smoothing can be used in the prediction of the rise and fall of stock prices. This helps the traders to have an idea of what to expect in the cost of trading.

Although smoothing reveals the patterns in a set of data, it provides no explanation as to why it is so. It is left to the researcher to draw conclusions as to the reasons for the patterns.

5 0

3 years ago

All of the following are suggestions of things to do if you have time left after completing the test except:

Fantom [35]

The correct option is B. Reread the directions for each section.

If you have time left after completing the test don't reread the directions for each section.

<h3>What is meant previewing a test?</h3>

In order to find the various problem kinds and their point values during the test preview, you must read the complete test. Mark the questions that you can answer quickly and easily.

The benefits of previewing the test are-

You will probably be given credit for the answers even if you accidentally copied them from the scratch paper.
Your effort on the scratch paper can earn you some points if a thoughtless mistake causes you to get the answer wrong.
If you do make a mistake, it will be simpler to find it when the instructor goes through the test.
By doing this, you can avoid making the same errors on your next exam.

Therefore, resolve each issue by re-entering the solution into the equation or performing the opposing operation needed to provide the appropriate response. Do not leave the testing room until the bell has rung or after you have gone over each problem twice.

To know more about Previewing, here

brainly.com/question/1144128

#SPJ4

The complete question is-

All of the following are suggestions of things to do if you have time left after completing the test except

A. Look at your answer sheet to make sure its filled properly

B. Reread the directions for each section

C. Return to the questioms you were unsure of

D. Make sure your answers correspond to the correct questions

6 0

2 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago