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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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Step-by-step explanation:

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Answer:

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b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

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Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

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The test statistic    z(e)  ??

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b) With value 51

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z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

8 0
3 years ago
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