Astudy of a company's practice regarding the payment of invoices revealed that an invoice was paid an average of 20 days after it was received. the standard deviation equaled five days. assuming that the distribution is normal, what percent of the invoices were paid within 15 days of receipt? multiple choice
a. 15.87%
b. 37.91%
c. 34.13%
d. 86.74%

=0.1587 or 15.87%

So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.

Step-by-step explanation:

In order to find the percent of the invoices paid within 5 days of receipt we have to find the value of Z first.

where:

X is the random varable which in our case is 15 days

u is the mean or average value which is 20 days

S is the standard deviation which is 5 days

Z=-1.0

We have to find Probability at Z less than -1

P(Z<-1.0) which can be written as:

=1-P(Z>1.0)

From Cumulative distribution table:

=1-(0.3413+0.5)

=0.1587 or 15.87%

So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}