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xxMikexx [17]
3 years ago
15

Find the LCM of n^3 t^2 and nt^4. A) n 4t^6 B) n 3t^6 C) n 3t^4 D) nt^2

Mathematics
2 answers:
patriot [66]3 years ago
8 0

Answer:

The correct option is C.

Step-by-step explanation:

The least common multiple (LCM) of any two numbers is the smallest number that they both divide evenly into.

The given terms are n^3t^2 and nt^4.

The factored form of each term is

n^3t^2=n\times n\times n\times t\times t

nt^4=n\times t\times t\times t\times t

To find the LCM of given numbers, multiply all factors of both terms and common factors of both terms are multiplied once.

LCM(n^3t^2,nt^4)=n\times n\times n\times t\times t\times t\times t

LCM(n^3t^2,nt^4)=n^3t^4

The LCM of given terms is n^3t^4. Therefore the correct option is C.

Sunny_sXe [5.5K]3 years ago
4 0

Answer: C) n 3t^4

Step-by-step explanation:

Definition : The least common multiple (LCM) of any two expressions is the smallest expression that is divisible by both expressions.

Given expressions : n^3 t^2 \text{ and } nt^4

Factorization form of n^3 t^2 \text{ and } nt^4 will be :

n^3 t^2 =n\times n\times n\times t\times t

tex]nt^4 =n \times t\times t\times t\times t[/tex]

The least common multiple of n^3 t^2 \text{ and } nt^4 :

n^3 t^2 =n\times n\times n\times t\times t\times t\times t=n^3t^4

Hence, the LCM of   n^3 t^2 \text{ and } nt^4=n^3t^4

Thus , the correct answer is option C).

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Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

The critical value of \frac{\alpha }{2} from the normal distribution table is  

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

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