2 POINTS) Using the calculation for damage and defects discussed in lecture, calculate the number of standardized sets that must be planned for in order to have 7,500 sets available to the customer each month. (Choose the closest answer) Group of answer choices 7431 7763 7604 7514 7798 7831 7708 7691 7512 7756

7431, 7763, 7604

Step-by-step explanation:

Number of standardized size Z

The observation 7500

Mean U

Standard deviation S

Group 1

7431 7763 7604

U = (7431 + 7763 + 7604)/4

U = -0.736

Z = (X - U)/S

Z = (7500 - 7599.3)/135 = -99.33/135

Z = - 0.736

For first group

Group 2

7514 7798 7831 7708

U = (7514+ 7798+ 7831 +7708 )/4

U = 7712.75

S = 123.27

Z = (X - U)/S

Z = (7500- 7712.75)/123.27 = -212.15/123.27

Z = - 1.721

For second group

Group 3

7691 7512 7756

U = (7691 + 7512 +7756)/3

U = 7652.7

S = 102.84

Z = (X - U)/S

Z = (7500- 7652.7)/102.84 = -152.7/102.84

Z = - 1.4848

For third group

Z1>Z3>Z2

The first group is the answer

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

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}