The lifetimes of lightbulbs of a particular type are normally distributed with a mean of 392 hours and a standard deviation of 9 hours. find the first quartile, q1.

The 1st quartile q1 is also called the 25th percentile (the bottom 25% of a data set). In this case, if we choose 100 light bulbs, then the 1st quartile will be the 25 bulbs with shortest useful life.

To solve this problem, we would have to use the z statistic. Using the standard distribution tables for z, we locate for the value of z at P = 0.25:

z = - 0.674

Then to calculate for the 1st quartile, we use the formula:

Q1 = x + z * s

where x is the mean, and s is the standard deviation, therefore:

Q1 = 392 + (- 0.674) (9)

Q1 = 385.934

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.

oof cheaters cheat because they’re lazy that’s why i cheat xd

: )

step-by-step explanation: