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3 years ago

Select the correct difference.

Mathematics

2 answers:

Mashutka [201]3 years ago

8 0

Answer:

+ 7 d - 7

7 d2 + d + 1

Step-by-step explanation:

+ 7 d - 7

7 d2 + d + 1

svetlana [45]3 years ago

3 0

+7d-7
7d2+d+1
Hope this helps

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Round 3.46 to the nearest integer

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Answer: 3.5 integers can be in decimals too. and any number counts as an integer except for 0

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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

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

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3 years ago

Hi! I need help solving these two problems. Can you help me please? They are attached below in the comments. Thank you!

AnnyKZ [126]

1. so the scale factor between the two shapes 2/3 because 6 is 2/3 of 9
and 9 is 3/2 of 6. Therefore y would equal 6 also, x would equal 8, because 8 is 2/3 of 12, and z would equal 5 1/4.
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3 years ago

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Answer:

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Step-by-step explanation:

$\frac{ {x}^{2} + {y}^{3} }{2 + x} \\$

when,

$x = 6 \\ y = 2$

Now substitute the value we get,

$\frac{ {6}^{2} + {2}^{3} }{2 + 6} \\ \\ = \frac{36 + 8}{8} \\ \\ = \frac{44}{8} \\ \\ = \frac{11}{2}$

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3 years ago

How to draw a number line to show that 2/3 is equivalent to 4/6

kondaur [170]

<-------------|-------------|-------------|------>
1/3 2/3 3/3

<------|------|------|------|------|------|------->
1/6 2/6 3/6 4/6 5/6 6/6

2/3 and 4/6 lines up the same :)

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