The weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a standard
deviation of 1 oz. What is the probability that a randomly
selected orange from the orchard weighs more than 4 oz., to the
nearest thousandth?

The probability is 0.933

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x-mean)/SD

x = 4

mean = 5.5

SD = 1

z-score = (4-5.5)/1 = -1.5

So we proceed to get the probability

This is;

P( z > -1.5)

we can get this from the standard normal distribution table

That will be

P(z > -1.5) = 0.93319

To the nearest thousandth, this is 0.933

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3