PLEASE HELP!!!!!!!!!

1 answer:

3 0

Answer: Choice D
z = 13/3 or z = -11/3

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Work Shown:

|z - 1/3| = 4
z - 1/3 = 4 or z - 1/3 = -4
using the rule that if |z| = k, then z = k or z = -k for some positive number k

Solve z - 1/3 = 4 to get...
z - 1/3 = 4
z - 1/3+1/3 = 4+1/3
z = 4+1/3
z = 12/3+1/3
z = (12+1)/3
z = 13/3

Solve z - 1/3 = -4 to get...
z - 1/3 = -4
z - 1/3+1/3 = -4+1/3
z = -4+1/3
z = -12/3+1/3
z = (-12+1)/3
z = -11/3

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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

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<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$