Weights of men: 90% confidence; n = 14, x=161.5 lb, s =13.7 lb

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 14

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.7709

The margin of error is:

M = T*s = 1.7709*13.7 = 24.26

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 161.5 - 24.26 = 137.24 lb

The upper end of the interval is the sample mean added to M. So it is 161.5 + 24.26 = 185.76 lb

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.

d

step-by-step explanation:

the solution is in the picture

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.