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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

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Answer:

Step-by-step explanation:

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3 years ago

Find the gradient of the line segment between the points (-4,-5) and (2,-4).

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Answer:

1/6

Step-by-step explanation:

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A company with a large fleet of cars wants to study the gasoline usage. They check the gasoline usage for 50 company trips chose

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Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

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And the interval would be (25.72; 28.32)

Step-by-step explanation:

Notation and definitions

n=50 represent the sample size

$\bar X= 27.2$ represent the sample mean

$s=5.83$ represent the sample standard deviation

m represent the margin of error

Confidence =88% or 0.88

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by $\alpha=1-0.88=0.12$ and $\alpha/2 =0.06$ . The degrees of freedom are given by: