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oee [108]
2 years ago
10

What is 0.7283 written in standard form​

Mathematics
2 answers:
Gnom [1K]2 years ago
5 0

Answer:

0.7283  written

in scientific is: 7.283×10^-1

Step-by-step explanation:

thats what i think but i'm not sure, hope you got it right sweetie!

Sav [38]2 years ago
5 0

Answer:

in scientific notation is 7.283×10^-1

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Space Shuttle has three computers. Computer A is a primary computer and Computer B and Computer C are auxiliary computers. There
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Answer:

Required probability equals 0.18%

Step-by-step explanation:

The probability that the primary and one auxiliary computer fails equals

1) Probability that A and B fails

2)Probability that A and C fails

Thus required probability equals

P(E)=P(1)(2)+P(1)P(3)\\\\P(E)=0.03\times 0.03+0.03\times 0.03P(E)=0.18%

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Solve the quadratic equation numerically (using tables of x- and y- values).
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Step-by-step explanation:

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Step-by-step explanation:

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

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

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

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

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
2 years ago
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