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3 years ago

Please answer this 18-34+49×200÷2

Mathematics

1 answer:

Kazeer [188]3 years ago

7 0

4884

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A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili

Fofino [41]

Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

$18x = 3*31$

$x = \frac{31*3}{18}$

$x = 5.17$

So $\mu = 5.17$

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

$P(X \leq 3) + P(X \geq 4) = 1$

We want $P(X \geq 4)$

$P(X \geq 4) = 1 - P(X \leq 3)$

In which

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057$

$P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294$

$P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760$

$P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242$

Finally

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758$

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

8 0

4 years ago

What step isolates the variable in the equaiton x/2 = 16? <br> (Will mark yo brainliest)

VashaNatasha [74]

Answer:

Multiply 2 i think!

Step-by-step explanation:

Hope this helped!

3 0

3 years ago

What is the value of r?<br><br> Enter your answer, as an exact value, in the box.

Oxana [17]

The answer is √32 because the triangle is an isosceles triangle meaning two sides are the same length so two sides are 4 and then you use the Pythagorean theorem to find the value of r

8 0

4 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

What is the solution of x - 9/ 7x + 2 < 0

vampirchik [111]

Answer:

x>7

Step-by-step explanation:

6 0

3 years ago

Please answer this 18-34+49×200÷2​

Please answer this 18-34+49×200÷2