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IgorLugansk [536]
3 years ago
5

The height of a cone-shaped statue is 9 ft, and the diameter is 12 ft. What is the volume of the statue? Use 3.14 to approximate

pi, and express your final answer to the nearest tenth. ft3   PLZ HELP
Mathematics
2 answers:
Anon25 [30]3 years ago
7 0

Answer:

≈ 339.1 ft^3

Step-by-step explanation:

Height = 9 ft

Radius = Diameter/2 = 12/2 = 6 ft

Volume of a cone = 1/3πr²h

= 2/3 × 3.14 × 6 × 6 × 9

= 339.12 ≈ 339.1 ft^3

neonofarm [45]3 years ago
4 0
Height = 9 ft
Radius = Diameter/2 = 12/2 = 6 ft
Volume of a cone = 1/3πr²h
= 2/3 × 3.14 × 6 × 6 × 9
= 339.12 ≈ 339.1 cubic foot.
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How is this solved..?
valina [46]

Answer:

Range : { -5,1,7}

Step-by-step explanation:

Take the values in the domain and substitute into the equation

x = -3

y = -2(-3) +1 = 6+1 =7

x = 0

y = -2(0) +1 = 0+1 =1

x = 3

y = -2(3) +1 = -6+1 =-5

The range is the y values

We put then in order from smallest to largest

Range : { -5,1,7}

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3 years ago
The mean and standard deviation of a random sample of n measurements are equal to and ​, respectively. a. Find a ​% confidence i
marusya05 [52]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

33.55 <  \mu < 35.5

b

34.03 <  \mu < 34.969

c

Generally the width at  n =  49 is mathematically represented as

   w =  2 *  E

    w =  2 *  0.952

     w =  1.904

Generally the width at  n =  196 is mathematically represented as

   w =  2 *  E

    w =  2 *  0.4687

     w =  0.9374

d

The correct option is E

Step-by-step explanation:

From the question we are told  that

   The sample mean is  \= x  =  34.5

    The standard deviation is  s =  3.4

Generally given that the confidence level is 95% then the level of significance is  

       \alpha = (100 -  95)\%

=>     \alpha = 0.05

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

   Z_{\frac{\alpha }{2} } =  1.96

Considering question a

From the question  n  =  49

Generally the margin of error is mathematically represented as  

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

=>   E = 1.96*  \frac{ 3.4 }{\sqrt{49} }

=>   E =  0.952

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  p <  \=x  +E

      34.5 -0.952 <  p < 34.5 + 0.952

=>    33.55 <  \mu < 35.5

Considering question b

From the question  n  =  196

Generally the margin of error is mathematically represented as  

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

=>   E = 1.96*  \frac{ 3.4 }{\sqrt{196} }

=>   E =  0.4687

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  p <  \=x  +E

      34.5 -0.4687 <  p < 34.5 +0.4687

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Considering question c

Generally the width at  n =  49 is mathematically represented as

   w =  2 *  E

    w =  2 *  0.952

     w =  1.904

Generally the width at  n =  196 is mathematically represented as

   w =  2 *  E

    w =  2 *  0.4687

     w =  0.9374

Now when the sample size is quadrupled i.e from n = 49 to  n =  196  

The width of the  confidence interval  decrease by 2 from 1.904 to  0.9374  

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Answer:

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