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Pachacha [2.7K]
3 years ago
10

Write the factored form of y2 + 11y + 28.

Mathematics
1 answer:
Anon25 [30]3 years ago
8 0

Answer:

\mathrm{Factor}\:y^2+11y+28:\quad \left(y+4\right)\left(y+7\right)

Step-by-step explanation:

Given the expression

y^2+11y+28

The expression can be written as into groups such as:

=\left(y^2+4y\right)+\left(7y+28\right)

\mathrm{Factor\:out\:}y\mathrm{\:from\:}y^2+4y\mathrm{:\quad }y\left(y+4\right)

\mathrm{Factor\:out\:}7\mathrm{\:from\:}7y+28\mathrm{:\quad }7\left(y+4\right)\\

so

=y\left(y+4\right)+7\left(y+4\right)

\mathrm{Factor\:out\:common\:term\:}y+4

=\left(y+4\right)\left(y+7\right)

Therefore,

\mathrm{Factor}\:y^2+11y+28:\quad \left(y+4\right)\left(y+7\right)

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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

c)

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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