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Juliette [100K]
2 years ago
11

A group of 59 randomly selected students have a mean score of 29.5 with a standard

Mathematics
1 answer:
algol [13]2 years ago
8 0

Answer:

29.5-1.671\frac{5.2}{\sqrt{59}}=28.37    

29.5+1.671\frac{5.2}{\sqrt{59}}=30.63    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

Step-by-step explanation:

Information given

\bar X = 29.5 represent the sample mean

\mu population mean  

s= 5.2 represent the sample standard deviation

n=59 represent the sample size  

Confidence interval

The confidence interval for the true mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom are given by:

df=n-1=59-1=58

The Confidence interval is 0.90 or 90%, the significance is \alpha=0.1 and \alpha/2 =0.05, and the critical value is t_{\alpha/2}=1.671

Now we have everything in order to replace into formula (1):

29.5-1.671\frac{5.2}{\sqrt{59}}=28.37    

29.5+1.671\frac{5.2}{\sqrt{59}}=30.63    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

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What is the solution to the system of equations?
elixir [45]

Answer:

No solution

Step-by-step explanation:

-3x-4y-3z=-7\\2x-6y+2z=3\\5x-2y+5z=9\\\\

Lets consider the equation 2.

Here we multiply the LHS and RHS with (-1).

-3x-4y-3z=-7\\-(2x-6y+2z)=-3\\5x-2y+5z=9\\\\

Hence,

-3x-4y-3z=-7\\-2x+6y-2z=-3\\5x-2y+5z=9\\\\

When we add the equations we get,

0x+0y+0z=-10+9\\Hence,\\0=-1

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6 0
2 years ago
A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3,
Romashka-Z-Leto [24]

Answer:

radius   x  = 3 ft

height   h = 23,8  ft

Step-by-step explanation:

From problem statement

V(t)  = V(cylinder) + V(hemisphere)

let x be radius of base of cylinder (at the same time radius of the hemisphere)

and h the height of the cylinder, then:

V(c)  = π*x²*h       area of cylinder = area of base + lateral area

                                              A(c)  = π*x²  +   2*π*x*h

V(h) = (2/3)*π*x³   area of hemisphere   A(h)  =   (2/3)*π*x²

A(t)  =  π*x²  +   2*π*x*h +   (2/3)*π*x²

Now A as fuction of x    

total volume   505  = π*x²*h  +  (2/3)*π*x³

h = [505 - (2/3)* π*x³ ]  /2* π*x    

Now we have the expression for A as function of x

A(x) =  3π*x² + 2π*x*h     A(x) = 3π*x² + 505  - (2/3)π*x³

Taking derivatives both sides

A´(x) =  6πx -  2πx²              A´(x) =  0         6x  - 2x²  = 0

x₁  =  0  we dismiss

6 - 2x = 0

x = 3     and   h = [505 - (2/3)* π*x³]/2* π*x

h  =  (505 - 18.84) / 6.28*3

h  = 23,8  ft

7 0
3 years ago
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