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Alex787 [66]
3 years ago
10

Is 832 is a perfect cube

Mathematics
2 answers:
Zepler [3.9K]3 years ago
4 0

Nope , 832 = 4×3√(13)

Alex777 [14]3 years ago
3 0

The Answer:

No sir it is not

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The average temperature for freezing foods is less than -17°C. Which of the following best describes this situation?
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D..............!! Idkkkkkkkkkkk
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A local food truck specializes in gourmet grilled cheese sandwiches. Each month, the owners of the truck have a constant total o
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B would be best answer. 5400s is rent

2.25s amount of sandwich cost and 9.00s selling prices. They all need s and has to be less then!
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Given that (8,-1) is on the graph of f(x), find
Nikolay [14]

Answer:

(-12, -1)

Step-by-step explanation:

(8, -1) => (-2/3x, f(-2/3x))

therefore

-2/3x = 8

x = 8 *-3/2

x = -12

f(-2/3x) = -1

therefore, the corresponding point is:

(-12,-1)

5 0
2 years ago
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
3 years ago
HURRY HELP
Gelneren [198K]
<h3>Answer: Choice (a)  $1,653.66</h3>

=====================================================

Work Shown:

Use the compound interest formula

A = P*(1+r/n)^(n*t)

A = 10000(1+0.039/1)^(1*4)

A = 11,653.65589441

A = 11,653.66

That's the amount of money Tammy will have at the end of 4 years.

Subtract off the principal from this value to get the interest.

interest = A - P

interest = 11,653.66 - 10,000

interest = 1,653.66

7 0
3 years ago
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