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Shtirlitz [24]
3 years ago
8

10 in 8 in 3 in 12 in Volume =

Mathematics
1 answer:
professor190 [17]3 years ago
4 0

Answer:

Your data is not accurate...

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Let P = 0.5 0.1 0.5 0.9 be the transition matrix for a Markov chain with two states. Let x0 = 0.5 0.5 be the initial state vecto
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Complete solution is given in the images attached below for better demonstration

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Urgent!!!! The scores of a high school entrance exam are approximately normally distributed with a given mean Mu = 82.4 and stan
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Step-by-step explanation:

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3 years ago
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Subtract 8 y^2 − 5 y + 7 from 2 y^2 + 7 y + 1 1 <br><br> The answer is: −6y ^2 +12y+4
lys-0071 [83]

Answer:

\large\boxed{-6y^2+12y+4}

Step-by-step explanation:

(2y^2+7y+11)-(8y^2-5y+7)\\\\=2y^2+7y+11-8y^2-(-5y)-7\\\\=2y^2+7y+11-8y^2+5y-7\qquad\text{combine like terms}\\\\=(2y^2-8y^2)+(7y+5y)+(11-7)\\\\=-6y^2+12y+4

4 0
3 years ago
Please help 10 point reward​
LekaFEV [45]

ANSWER:    I can give you the steps to figure out this question.

1  if it is starting at the 6 on the y axis, then just do up 1 over 1 until you get to the 10

The answer to your question is..... 6y+1=4x+10

4 0
3 years ago
In a clothing store, 65% of the customers buy a shirt, 30% of the customers
diamong [38]

Answer:

75%

Step-by-step explanation:

List out known probabilities

\Pr(shirt) = 0.65 = \frac{65}{100} = \frac{13}{20}\\ \Pr(pants) = 0.40 = \frac{30}{100} = \frac{6}{20}\\\Pr(shirt \cap pants) = 0.20 = \frac{20}{100}=\frac{1}{5}

recall equation

\Pr(A \cup B)=\Pr(A)+\Pr(B)-\Pr(A \cap B)

Plug in values

\Pr(shirt \cup pants)=\frac{13}{20}+\frac{6}{20}-\frac{1}{5}\\\\therefore \Pr(shirt \cup pants) = \frac{15}{20} =\frac{3}{4} = 75\%

4 0
2 years ago
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