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asambeis [7]
4 years ago
7

The upper quartile for the data set given below is ------- 14, 8, 23, 9, 11, 27, 22, 3, 17, 12, 29

Mathematics
2 answers:
Darya [45]4 years ago
4 0
The answer is 22.5... 22+23=45...45/2=22.5
aleksklad [387]4 years ago
3 0

Answer:

the answer is 23

Step-by-step explanation:

i know it

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It is greatest to least fractions
Lana71 [14]

Answer:

D

Step-by-step explanation:

The best way to work with fractions is to turn them into decimals by dividing the numerator by the denominator.

7 0
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

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

4 0
3 years ago
WILL GIVE BRANLIEST AND BE SOOO HAPPY PLEASE HELP!!! 30 POINTS SHARE YOUR SMARTNESS!!
OlgaM077 [116]

Step-by-step explanation:

For the 1st picture,

The 2nd option is the answer.

Let say we have

a_{3}

Replace it into the recursive formula.

a_{3} =  a_{3 - 2} + (a)   _{3 - 1}

We are given a,1 and a,2 so the 3rd terms has to be

2

So the second option is the answer.

For the second picture, a geometric sequence is a group of terms that has the same ratio between each consecutively.

Infinite means going on forever

The fourth option is the answer.

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3 years ago
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Karo-lina-s [1.5K]
12 girls for every 5 boys
boys is second so it shouldb be indenomenaotr (bottom)

girls/boys

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6 0
3 years ago
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What is the value of b?<br> A 76 <br> B 106 <br> C 120 <br> D 224
fredd [130]
68 + a = 180
a = 180 - 68
a = 112
b = 2(112) - 104
b = 224 - 104
b = 120

answer
<span>C 120 </span>
6 0
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