The weights of packages of crackers are normally distributed with a mean of 340 grams and a standard deviation of 12 grams.

What is the weight of a package of crackers with a z-score of −1.4 ?

Enter your answer in the box.

the weight of the package of crackers is 323.2 grams

Step-by-step explanation:

Since the weights of packages of crackers are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = weights of packages of crackers.

µ = mean weight

σ = standard deviation

From the information given,

µ = 340 grams

σ = 12 grams

The z score of the package is - 1.4

Therefore,

- 1.4 = (x - 340)/12

Cross multiplying by 12, it becomes

- 1.4 × 12 = x - 340

- 16.8 = x - 340

x = - 16.8 + 340

x = 323.2 grams

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

fwms@lil_ben06

step-by-step explanation: