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Anastaziya [24]
3 years ago
13

How do I solve this equation? 3|13-2t|=15

Mathematics
2 answers:
Sholpan [36]3 years ago
7 0
/ 13 - 2t / = 15 ÷ 3 ;
/ 13 - 2t / = 5 ;
13 - 2t = + 5 or 13 - 2t = -5 ;
First equation, -2t = - 8; t = 4 ;
Second equation, -2t = - 18 ; t = 9 ;
The solutions are 4 and 9 .
Nikolay [14]3 years ago
5 0
3 |13 - 2t | =15 ⇒ |13-2t| = 15\div3 ⇒ \boxed {|13-2t| = 5}

\hbox {We have two equations : 1. } |13-2t|=5 
                                               2.|13-2t|=-5

\hbox{The first: } ⇒ 13-2t=5 ⇒ 2t=13-5 ⇒ 2t=8 ⇒ \boxed {t=4}

\hbox { The second: } 13-2t=-5 ⇒ 2t=13-(-5)=18 ⇒ \boxed {t=9}
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b) 0.5065 = 50.65%.

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<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
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For this problem, the fixed parameter is:

p = 0.37.

Item a:

The probability is P(X = 1) when n = 1, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{1,1}.(0.37)^{1}.(0.63)^{0} = 0.37

Item b:

The probability is P(X = 3) when n = 3, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.37)^{3}.(0.63)^{0} = 0.5065

Item c:

The probability is P(X = 2) when n = 4, hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(0.37)^{2}.(0.63)^{2} = 0.3260

More can be learned about the binomial distribution at brainly.com/question/24863377

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