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user100 [1]
3 years ago
6

A recent study of 26 city residents showed that the time they had lived at their present address has a mean of 10.3 years with t

he standard deviation of 3 years. Find the 95% confidence interval of the true mean. Assume that the variable is approximately normally distributed.
Mathematics
1 answer:
svetlana [45]3 years ago
7 0

Answer:

The 95% confidence interval is  9.15<  \mu < 11.45

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  26

      The mean is  \= x =  10.3

       The standard deviation is  s =  3

From the question we are told the confidence level is  95% , hence 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

Generally the margin of error is mathematically represented as  

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

=>   E =  1.96 *  \frac{3}{\sqrt{26} }

=>   E = 1.1532

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>   10.3  -1.1532 <  \mu < 10.3  +  1.1532

=>    9.15<  \mu < 11.45

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zheka24 [161]

Answer:

15.4%

Step-by-step explanation:

Calculate the volume of both containers.

The volume (V) of a cylinder is calculated as

V = πr²h ( r is the radius and h the height )

V_{A} = π × 10² × 11 = 1100π ft³

V_{B} = π × 7² × 19 = 931π ft³

Subtract volume of B from volume of A to find quantity left in A

quantity left in A = 1100π - 931π = 169π ft³

To calculate the percentage left in A after B has been filled

\frac{left}{full} × 100%

= \frac{169\pi }{1100\pi } × 100%

= \frac{169}{1100} × 100%

= 15.4% ( to the nearest tenth )

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Given: DR tangent to Circle O.<br> If m _ RDC = 120°, then m DAC =<br> •60<br> •120<br> •240
Katyanochek1 [597]

Answer:

The measure of the arc DAC is 240°

Step-by-step explanation:

we know that

The semi-inscribed angle is half that of the arc it comprises.

so

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we have

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substitute

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Rewrite

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7 0
3 years ago
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Find a degree 3 polynomial with real coefficients having zeros 2 and 2 − 2 i and a lead coefficient of 1. Write P in expanded fo
Zanzabum

Answer:

The polynomial with real coefficients having zeros 2 and 2 - 2i is

x³ - 6x² + 16x - 16 = 0

Step-by-step explanation:

Given that a polynomial has zeros at 2 and 2 - 2i, we want to write this polynomial.

We have

x - 2 = 0

x - (2 - 2i) = 0

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Since the polynomial has real coefficients, and 2 - 2i is a zero of the polynomial, the conjugate of 2 - 2i, which is 2 + 2i is also a polynomial.

x - (2 + 2i) = 0

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Now,

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= (x - 2)((x - 2)² - (2i)²) = 0

= (x - 2)(x² - 4x + 8) = 0

= x³ - 4x² + 8x - 2x² + 8x - 16 = 0

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This is the polynomial required.

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(5x^3)(4x)^3
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3 years ago
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