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.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

bull

step-by-step explanation: