Urrently quarters have weights that are normally distributed with a mean of 5.670 g and a standard deviation of 0.062 g a vending machine is configured to accept only those quarters with weights between 5.550g if 280 diffferent quarters are inserted into the vending machine what is the probability that the mean falls between the limits of 5.550g and 5.790g a. 0.0002 b.0.9476 c.0.9998 d.0.0524

B) .9476

Step-by-step explanation

Its easiest to solve on a TI calculator.

1. Identify your mean, standard deviation, and your lower/high limits. (Lower = 5.550 Higher = 5.790)

2. Now use calculator..

2nd>vars>normalcdf

Fill it in with your data then click paste.

(IF YOU HAVE TO SHOW YOUR WORK WRITE OUT WHATS GIVEN ON YOUR CALC..  normalcdf(5.550,5.790,5.670,.062) )

= .9470691924

Rounded to four decimals..

none.

step-by-step explanation:

m

1

=

5

m1=5

b

=

−

1

b=-1

find the slope and y-intercept of the second equation.

tap for more

m

2

=

0

m2=0

b

=

−

6

b=-6

compare the slopes

m

m of the two equations.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

compare the two slopes in the decimal form. if the slopes are equal, then the lines are parallel. if the slopes are not equal, then the lines are not parallel.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

0lz try yourself sorry for not answering