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3 years ago

11.2x-6.3=39.2 Please help

Mathematics

1 answer:

Ilya [14]3 years ago

5 0

<h3><u>Add 6.3 to both sides:</u></h3>

2x-6.3=39.2

2x-6.3+6.3=39.2+6.3

<h3><u>simplify:</u></h3>

2x=45.5

<h3><u>Divide both sides of the equation by the same term:</u></h3>

2x=45.5

2x/2 = 45.5/2

<h3><u>Simplify:</u></h3>

x=22.75

<h3><u>Answer:</u></h3>

x=22.75

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3 years ago

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

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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

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3 years ago