Q4 - Exam prep: Missing values Audrey measures all 7 pencils in her pencil case and writes the first 6 lengths down.
12.4 cm, 12.6 cm, 13 cm, 13.4 cm, 13.5 cm, 13.3 cm
Audrey calculates the mean length of all 7 pencils to be 13.1 cm.
Find the length of the missing pencil.
cm
[2]

the missing pencil length is 13.5 cm

Step-by-step explanation:

The computation of the length of the missing pencil is shown below:

Let us assume the missing pencil be x

Now

(12.4 cm + 12.6 cm + 13 cm + 13.4 cm + 13.5 cm + 13.3 cm + x) ÷ 7 = 13.1 cm

(78.2 cm + x) ÷ 7 = 13.1 cm

78.2 cm + x = 91.7 m

So, x would be

= 91.7 cm - 78.2 cm

= 13.5 cm

hence, the missing pencil length is 13.5 cm

the correct answer to your question is a. 4^4

good luck my friend!

hope this !

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! < 3

have a splendid day < 3 : )

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}