All categories

2 years ago

What is the distance between [(3 + 4i) + (2 - 3i)] and (9 - 2i)?

Mathematics

1 answer:

Semmy [17]2 years ago

5 0

Answer:

Step-by-step explanation:

(3 + 4i) + (2 - 3i) = 3 + 4i + 2 - 3i = 5 + i

distance between (5 + i) and (9 - 2i) is the difference between them. and difference means subtraction.

(9 - 2i) - (5 + i) = 9 - 2i - 5 - i = 4 - 3i

and since we are looking for a distance, we are looking for the absolute value of that subtraction.

after all, we could have done the subtraction also in the other direction

(5 + i) - (9 - 2i) = -4 + 3i

and this must be the same distance.

|(-4 + 3i)| = |(4 - 3i)|

and that is done by calculating the distance of any of these 2 points from (0,0) on the coordinate grid of complex numbers.

|(a +bi)| = sqrt(a² + b²)

in our case here

distance = sqrt(4² + (-3)²) = sqrt(16 + 9) = sqrt(25) = 5

as you can easily see, this is (as expected) the same for the result of the subtraction in the other direction :

sqrt((-4)² + 3²) = sqrt(16+9) = sqrt(25) = 5

You might be interested in

Tina has 6 packs of 8 colored markers to put into party gift bags she spirits markers equally among 9 gift bags. How many marker

lyudmila [28]

Answer:

252

Step-by-step explanation:

8 0

3 years ago

FInd mPQ <br> Geometry question, circles, arc length, chord length,

Fed [463]

Answer:

$\text{B. }26}$

Step-by-step explanation:

Since $\overline{PN}$ and $\overline{PQ}$ are equal, we can set them equal to each other:

$7x+5=10x-4,\\9=3x,\\x=3$

Substitute $x=3$ to solve for $\overline{PQ}$ :

$\overline{PQ}=10(3)-4=\boxed{\text{B. }26}}$

3 0

3 years ago

With complete solution please I need Thanks

Liula [17]

Answer:

Step-by-step explanation:

We can make this and equation

x • y = 1260

x = 3y +48

Our equation is

(3y +48)y = 1260

We get y = 14

6 0

2 years ago

What is 7 times 7 + 7 ?

Basile [38]

The answer would be 98

7+7=14
14•7=98

5 0

2 years ago

Read 2 more answers

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

Since is a one-side upper test the p value would be:

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

First we need to calculate the degrees of freedom given by:

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

Since is a one-side upper test the p value would be:

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

3 years ago