Data on investments in the high-tech industry by venture capitalists are compiled by a corporation. A random sample of 18venture-capital investments in a certain business sector yielded the accompanying data, in millions of dollars. Determine and interpret a 95%confidence interval for the mean amount, mu, of all venture-capital investments in this business sector. Assume that the population standard deviation is $1.70million. (Note: The sum of the data is $102.52million.)

$4.911  million or  $6.481 million

Thus, we are 95% confident that the mean amount of all venture-capital investments in the high-tech industry is somewhere between $4.911 million and $6.481 million.

Step-by-step explanation:

Given that:

sample size n = 18

standard deviation σ = 1.70

confidence interval = 95%

Sample mean

5.696

The level of significance = 1 - C.I

= 1 - 0.95

= 0.05

The critical value of from the Z tables

The 95% C.I for the mean is;

= 5.696 ± 0.785372

= (5.696 - 0.785372 , 5.696 + 0.785372 )

= ( 4.910628 , 6.481372  )

≅ $4.911  million or  $6.481 million.

Thus, we are 95% confident that the mean amount of all venture-capital investments in the high-tech industry is somewhere between $4.911 million and $6.481 million.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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