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2 years ago

If z1 =2cis120, z2=4cis20, and z1/z2=a+bi, then a = and b = ?

Mathematics

1 answer:

balu736 [363]2 years ago

5 0

Hello from MrBillDoesMath!

Answer:

a = 0, b = 1/2

Discussion:

z1 = 2 cis120, z2 = 4 cis30

NOTE: The question states z2 = 4cis20 but the image shows 4cis30. I'm using the values shown in the image.

z1/z2 =

2 cis120 / 4 cis30 =

(2/4) cis(120-30) =

(1/2) cis(90) =

(1/2) (cos90 + i sin90) = => as cos90 = 0 and sin90 =1

(1/2) (0 + i) =

0 + (1/2) i

If z1/z2 = a + bi, then a = 0. and b = 1/2

Thank you,

MrB

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a) We have to perform a hypothesis test for the difference of proportions.

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$H_0: \pi_1\geq\pi_2\\\\H_1: \pi_1$

The significance level is 0.05.

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$p_2=\frac{54}{399}=0.1353$

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The estimated standard deviation is

$s=\sqrt{\frac{p(1-p)}{n_1}+\frac{p(1-p)}{n_2}} =\sqrt{\frac{0.1153(1-0.1153)}{2142}+\frac{0.1153(1-0.1153)}{399}} =0.0174$

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b) We can construct a 95% CI, according to the significance level of 0.05.

The z-value for this CI is 1.96.

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$\sigma=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2}} =\sqrt{\frac{0.1116(1-0.1116)}{2142} +\frac{0.1353(1-0.1353)}{399}} =0.0184$

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$LL=(p_1-p_2)-z*\sigma=(0.1116-0.1353)-1.96*0.0184=-0.0238-0.0361\\\\LL=-0.0599$

The upper limit is:

$UL=(p_1-p_2)+z*\sigma=(0.1116-0.1353)+1.96*0.0184=-0.0238+0.0361\\\\UL=0.0124$

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$-0.0599\leq\pi_1-\pi_2\leq0.0124$

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2 years ago

If z1 =2cis120, z2=4cis20, and z1/z2=a+bi, then a = __ and b = __?

If z1 =2cis120, z2=4cis20, and z1/z2=a+bi, then a = and b = ?