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amid [387]
3 years ago
12

How many ways can six of the letters of the word ALGORITHM be selected 8. How many ways can the letters of the word ALGORITHM be

arranged in a be seated together in the row? and written in a row? row if the letters GOR must remain together (in this order)?
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
3 0
<h2>Answer with explanation:</h2>

The number of letters in word "ALGORITHM" = 9

The number of combinations to select r things from n things is given by :-

C(n,r)=\dfrac{n!}{r!(n-r)!}

Now, the number of combinations to select 6 letters from 9 letters will be :-

C(9,8)=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!\times3!}=84

Thus , the number of ways can six of the letters of the word ALGORITHM=84

The number of ways to arrange n things in a row :n!

So, the number of ways can the letters of the word ALGORITHM be arranged in a be seated together in the row :-

9!=362880

If GOR comes together, then we consider it as one letter,  then the total number of letters will be = 1+6=7

Number of ways to arrange 9 letters if "GOR" comes together :-

7!=5040

Thus, the number of ways to arrange 9 letters if "GOR" comes together=5040

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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri
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Answer:

a

 P(\= X \ge 51 ) =0.0062

b

P(\= X \ge 51 ) = 0

Step-by-step explanation:

From the question we are told that

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The standard deviation is  \sigma = 1.2

Considering question a

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Generally the standard error of the mean is mathematically represented as

      \sigma_x = \frac{\sigma }{\sqrt{n} }

=>   \sigma_x = \frac{ 1.2 }{\sqrt{9} }

=>  \sigma_x = 0.4

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

      P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}}  \ge \frac{51 - 50 }{0.4 } )

\frac{\= X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ \= X )

     P(\= X \ge 51 ) = P( Z  \ge 2.5 )

=>   P(\= X \ge 51 ) =1-  P( Z  < 2.5 )

From the z table  the area under the normal curve to the left corresponding to  2.5  is

    P( Z  < 2.5 ) = 0.99379

=> P(\= X \ge 51 ) =1-0.99379

=> P(\= X \ge 51 ) =0.0062

Considering question b

The sample size is  n = 40

   Generally the standard error of the mean is mathematically represented as

      \sigma_x = \frac{\sigma }{\sqrt{n} }

=>   \sigma_x = \frac{ 1.2 }{\sqrt{40} }

=>  \sigma_x = 0.1897

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as  

       P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x}  \ge \frac{51 - 50 }{0.1897 } )

=> P(\= X \ge 51 ) = P(Z  \ge 5.2715  )

=>  P(\= X \ge 51 ) = 1- P(Z < 5.2715  )

From the z table  the area under the normal curve to the left corresponding to  5.2715 and

=>  P(Z < 5.2715  ) = 1

So

   P(\= X \ge 51 ) = 1- 1

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Answer:

t_1\approx2.18

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0=-16t^2+44t-20

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t=\frac{-44\pm\sqrt{44^2-4(-16)(-20)}}{2(-16)}

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