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grandymaker [24]

4 years ago

15

Complex numbers, simplify the following expression; 10/1+3i

1 answer:

svp [43]4 years ago

8 0

$\frac{10}{1 + 3i} = \frac{10}{1 + 3i} * \frac{1-3i}{1-3i} = \frac{10-30i}{1^{2} -(3i)^{2}} = \frac{10-30i}{1 -9 i^{2} } = \frac{10-30i}{1 -9*(-1)} = \frac{10(1-3i)}{1+9}= \frac{10(1-3i)}{10}$ $= 1 -3i$

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Olenka [21]

Answer:

a

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From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

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Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

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=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

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Fynjy0 [20]

Answer:

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