The lifetimes of a certain type of calculator battery are normally distributed. The mean lifetime is 400 days, with a standard deviation of 50 days. For a sample of 6000 new batteries, determine how many batteries will last:

Complete question :

The lifetimes of a certain type of calculator battery are normally distributed. The mean lifetime is 400 days, with a standard deviation of 50 days. For a sample of 6000 new batteries, determine how many batteries will last: 360 and 460 days

0.67307

Step-by-step explanation:

Given that :

Mean, m = 400

Standard deviation, s = 50

Sample size, n = 6000

Obtain the standardized score :

Zscore =(x - m) / s

For X = 360

P(x < 360)

Zscore =(360 - 400) / 50

Zscore = - 40 / 50

Zscore = - 0.8

P(Z < - 0.8) = 0.21186

For X = 460

P(x < 460)

Zscore =(460 - 400) / 50

Zscore = 60 / 50

Zscore = 1.2

P(Z < 1.2) = 0.88493

P(Z < 1.2) - P(Z < - 0.8)

0.88493 - 0.21186

= 0.67307

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²