The life expectancy (in hours) of an electric bulb is normally distributed with a mean of 5000 and a standard deviation of 1000. find the probability that a bulb lasts for more than 6300 hours. round answer to four decimal places. (tip: calculate z-value and use z-probability distribution table.)

P(z>1.3) = 0.9032

Step-by-step explanation:

We are given:

Mean = 5000

Standard deviation = 1000

x = 6300

P(x>6300)=?

z-score =?

z-score = x- mean/standard deviation

z-score = 6300 - 5000/1000

z- score = 1300/1000

z-score = 1.3

So, P(x>6300) = P(z>1.3)

Looking at the z-probability distribution table and finding value:

P(z>1.3) = 0.9032

So, P(z>1.3) = 0.9032

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.

you can not solvr this without numbers