(1) If random variable X follows the Binomial distribution X~Binomi.pdf

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The document contains examples and explanations of probability calculations involving the binomial distribution. It shows how to calculate the probability of outcomes being less than, equal to, or greater than certain values. For example, it demonstrates calculating the probability of a random variable X being greater than 4, given X follows a binomial distribution with n=12 and p=0.5. It also provides the formula for calculating probabilities within a certain range.

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(1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the
following statements is/are correct?
P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6)
False
P(X<6) = P(0) + P(1) + ... + P(5)
** You do NOT include 6.
P(X<=20)=1
True
Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0.
P(X< = 8) + P(9 < = X < = 20)
= 1 + 0
= 1
X can take 8 different values.
False
X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8
P(X>=3)+P(X<=3)=1
False
P(X > = 3) + P(X < = 2) = 1
P(X>6)=1-P(X<7)
True
P(X > 6)
= P(X > = 7)
= 1 - P(X < = 6)
= 1 - P(X < 7)
P(2
Unknown - this statement is incomplete
---------------

Note: The following notation varies from book-to-book.
Given: n = 12 and p = .5
P(X) = nCx * p^x * (1-p)^(n-x)
P(X = 5) = 12C5 * .5^5 * .5^7
P(X = 5) = 0.19336
(b)
P(X > 4)
= P(5) + P(6) + ... + P(12) <-- 8 probabilities to find
= 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find
= 8C4 * 0.5^4 * 0.5^8 = 0.12085
+ 8C3 * 0.5^3 * 0.5^9 = 0.05371
+ 8C2 * 0.5^2 * 0.5^10 = 0.01611
+ 8C1 * 0.5^1 * 0.5^11 = 0.00293
+ 8C0 * 0.5^0 * 0.5^12 = 0.00024
= 1 - [0.12085 + 0.05371 + ... + 0.00024]
= 0.806
---------------
(1)
Given: n = 20 and p = .01
(a)
P(X = 0)
= 20C0 * 0.01^0 * 0.99^20 = 0.818
(b)
P(at most 1) = P(1) + P(0)
= 20C1 * 0.01^1 * 0.99^19 = 0.16523
+ 20C0 * 0.01^0 * 0.99^20 = 0.81791
= 0.983

(c)
P(more than 1) = P(2) + P(3) + ... + P(20)
= 1 - [P(1) + P(0)]
= 1 - 0.983
= 0.017
---------------
(1)
Given: n = 20 and p = .01
P(X = 0)
= 20C0 * 0.01^0 * 0.99^20 = 0.818
Given: n = 20 and p = .02
P(X = 0)
= 20C0 * 0.02^0 * 0.98^20 = 0.668
Given: n = 20 and p = .05
P(X = 0)
= 20C0 * 0.05^0 * 0.95^20 = 0.358
---------------
(1)
(a) mean = (6+2)/2 = 4
(b)
P(X > 4)
= P(4 < X < 6)
= (6-4)/(6-2)
= 1/2
= .5

Note: P(a < X < b) = (b - a)/(Max - Min)
(c)
= P(3 < X < 5)
= (5-3)/(6-2)
= 1/2
= .5
---------------
I hope this helped. If you have any questions, please ask them in the comment section. :)
Solution
(1) If random variable X follows the Binomial distribution: X~Binomial(8,0.7), which among the
following statements is/are correct?
P(X<6)=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5)+p(x=6)
False
P(X<6) = P(0) + P(1) + ... + P(5)
** You do NOT include 6.
P(X<=20)=1
True
Since n = 8, X can't be 9 or larger, so P(9 < = X < = 20) = 0.
P(X< = 8) + P(9 < = X < = 20)
= 1 + 0
= 1
X can take 8 different values.
False
X can take on 9 values: 0, 1, 2, 3, 4, 5, 6, 7 and 8
P(X>=3)+P(X<=3)=1
False

P(X > = 3) + P(X < = 2) = 1
P(X>6)=1-P(X<7)
True
P(X > 6)
= P(X > = 7)
= 1 - P(X < = 6)
= 1 - P(X < 7)
P(2
Unknown - this statement is incomplete
---------------
Note: The following notation varies from book-to-book.
Given: n = 12 and p = .5
P(X) = nCx * p^x * (1-p)^(n-x)
P(X = 5) = 12C5 * .5^5 * .5^7
P(X = 5) = 0.19336
(b)
P(X > 4)
= P(5) + P(6) + ... + P(12) <-- 8 probabilities to find
= 1 - [P(4) + P(3) + ... + P(0)] <-- 5 probabilities to find
= 8C4 * 0.5^4 * 0.5^8 = 0.12085
+ 8C3 * 0.5^3 * 0.5^9 = 0.05371
+ 8C2 * 0.5^2 * 0.5^10 = 0.01611
+ 8C1 * 0.5^1 * 0.5^11 = 0.00293
+ 8C0 * 0.5^0 * 0.5^12 = 0.00024
= 1 - [0.12085 + 0.05371 + ... + 0.00024]
= 0.806
---------------

(1)
Given: n = 20 and p = .01
(a)
P(X = 0)
= 20C0 * 0.01^0 * 0.99^20 = 0.818
(b)
P(at most 1) = P(1) + P(0)
= 20C1 * 0.01^1 * 0.99^19 = 0.16523
+ 20C0 * 0.01^0 * 0.99^20 = 0.81791
= 0.983
(c)
P(more than 1) = P(2) + P(3) + ... + P(20)
= 1 - [P(1) + P(0)]
= 1 - 0.983
= 0.017
---------------
(1)
Given: n = 20 and p = .01
P(X = 0)
= 20C0 * 0.01^0 * 0.99^20 = 0.818
Given: n = 20 and p = .02
P(X = 0)
= 20C0 * 0.02^0 * 0.98^20 = 0.668
Given: n = 20 and p = .05
P(X = 0)
= 20C0 * 0.05^0 * 0.95^20 = 0.358

---------------
(1)
(a) mean = (6+2)/2 = 4
(b)
P(X > 4)
= P(4 < X < 6)
= (6-4)/(6-2)
= 1/2
= .5
Note: P(a < X < b) = (b - a)/(Max - Min)
(c)
= P(3 < X < 5)
= (5-3)/(6-2)
= 1/2
= .5
---------------
I hope this helped. If you have any questions, please ask them in the comment section. :)

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PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150 T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267 Solution PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150 T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267.

PV ^ n = constant using pv = N R T we get P ^ .pdf

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methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a higher level of polarity, and the compound containing the nitro is more polar. Hope this helps! Solution methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a higher level of polarity, and the compound containing the nitro is more polar. Hope this helps!.

methyl-m-nitrobenzoate is more polar than methyl .pdf

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L(f(t)) = 6s^4 - 2s^3 .pdf

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It was a 19th century geologist who first recognized periodicity in the physical properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The list was constructed as a helical graph wrapped around a cylinder--elements with similar properties occupied positions on the same vertical line of cylinder (the list also included some ions and compounds). Using geological terms and published without the diagram, de Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to distinguish elements by their properties Solution It was a 19th century geologist who first recognized periodicity in the physical properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The list was constructed as a helical graph wrapped around a cylinder--elements with similar properties occupied positions on the same vertical line of cylinder (the list also included some ions and compounds). Using geological terms and published without the diagram, de Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to distinguish elements by their properties.

It was a 19th century geologist who first recogni.pdf

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