All categories

3 years ago

What is i^84? a) -i b) i c) -1 d) 1

Mathematics

1 answer:

mariarad [96]3 years ago

7 0

Answer:

D) 1

Step-by-step explanation:

When you start raising i to certain powers, you begin to notice a pattern.

$i^1=i \\\\i^2=-1 \\\\i^3=-i \\\\i^4=1 \\\\i^5=i \\\\ i^6=-1 \\\\ i^7=-i \\\\i^8= 1$

This cycle repeats forever. Since 84 is a multiple of 4, i^84 must be 1. Hope this helps!

You might be interested in

Last year, Deshaun had $20,000 to invest. He invested some of it in an account that paid 9% simple interest per year, and he inv

dezoksy [38]

Answer:

the amount Deshaun invested in an account that paid 9% interest = $9,000

the amount Deshaun invested in an account that paid 5% interest = $11,000

Step-by-step explanation:

Let x = the amount Deshaun invested in an account that paid 9% interest

Let y = the amount Deshaun invested in an account that paid 5% interest.

Thus;

x + y = 20,000 - - - - (eq 1)

Now, After one year, he received a total of $1440 in interest. Thus;

0.09x + 0.05y = 1440

Multiply each term by 100 to give;

9x + 5y = 144000 ---- (eq 2)

Making x the subject in equation 1,we have;

x = 20,000 - y

Putting 20,000 - y for x in eq 2,we have;

9(20,000 - y) + 5y = 1440

180000 - 9y + 5y = 144000

180000 - 4y = 144000

180000 - 144000 = 4y

36000 = 4y

Divide both sides by 4 to give;

y = 36000/4

y = $9,000

x = 20,000 - 9000

x = $11,000

7 0

3 years ago

Round 74952 to the nearest 1000

stellarik [79]

75,000 is the answer!

If this was the appropiate answer make sure to mark as the brainliest!
-procklown

6 0

3 years ago

Read 2 more answers

A professor at a local community college noted that the grades of his students were normally distributed with a mean of 84 and a

creativ13 [48]

Answer:

A. P(x>91.71)=0.10, so the minimum grade is 91.71

B. P(x<72.24)=0.025 so the maximum grade could be 72.24

C. By rule of three, 200 students took the course

Step-by-step explanation:

The problem says that the grades are normally distributed with mean 84 and STD 6, and we are asked some probabilities. We can´t find those probabilities directly only knowing the mean and STD (In that distribution), At first we need to transfer our problem to a Standard Normal Distribution and there is where we find those probabilities. We can do this by a process called "normalize".

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

Where x,a are data from the original normal distribution, μ is the mean, σ is the STD and z,b are data in the Standard Normal Distribution.

There´s almost no tools to calculate probabilities in other normal distributions. My favorite tool to find probabilities in a Standard Normal Distribution is a chart (attached to this answer) that works like this:

P(x<c=a.bd)=(a.b , d)

Where "a.b" are the whole part and the first decimal of "c" and "d" the second decimal of "c", (a.b,d) are the coordinates of the result in the table, we will be using this to answer these questions. Notice the table only works with the probability under a value (P(z>b) is not directly shown by the chart)

A. We are asked for the minimum value needed to make an "A", in other words, which value "a" give us the following:

P(x>a)=0.10

Knowing that 10% of the students are above that grade "a"

What we are doing to solve it, as I said before, is to transfer information from a Standard Normal Distribution to the distribution we are talking about. We are going to look for a value "b" that gives us 0.10, and then we "normalize backwards".

P(x>b)=0.10

Thus the chart only works with probabilities UNDER a value, we need to use this property of probabilities to help us out:

P(x>b)=1 - P(x<b)=0.10

P(x<b)=0.9

And now, we are able to look "b" in the chart.

P(x<1.28)=0.8997

If we take b=1.285

P(x<1.285)≈0.9

Then

P(x>1.285)≈0.1

Now that we know the value that works in the Standard Normal Distribution, we "normalize backwards" as follows:

P(x<a) = P( (x-μ)/σ < (a-μ)/σ ) = P(z<b)

If we take b=(a+μ)/σ, then a=σb+μ.

a=6(1.285)+84

a=91.71

And because P(x<a)=P(z<b), we have P(x>a)=P(z>b), and our answer will be 91.71 because:

P(x>91.71) = 0.1

B. We use the same trick looking for a value in the Standard Normal Distribution that gives us the probability that we want and then we "normalize backwards"

The maximum score among the students who failed, would be the value that fills:

P(x<a)=0.025

because those who failed were the 2.5% and they were under the grade "a".

We look for a value that gives us:

P(z<b)=0.025 (in the Standard Normal Distribution)

P(z<-1.96)=0.025

And now, we do the same as before

a=bσ+μ

a=6(-1.96)+84

a=72.24

So, we conclude that the maximum grade is 72.24 because

P(x<72.24)=0.025