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3 years ago

Which sequence explains a geometric method of determining 3 - 17i/2 +2i?

Mathematics

2 answers:

Katen [24]3 years ago

8 0

Answer:

D) Plot 3 – 17i. Scale by 2sqrt(2). Rotate 45° clockwise.

Step-by-step explanation:

never [62]3 years ago

7 0

Answer:

That would be D.

Step-by-step explanation:

first, turn both to polar form. Then you plot the first one and scale by the second. also, using a Rectangular to Polar calculator, we can see that 2.82 is 2 sqrt 2, so it cannot be A or B. It also can't be C since you graph 3-17i first.

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PLZ HELP!!!!!!!!! IM BEING TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mars2501 [29]

Answer:

B. 40

Step-by-step explanation:

It looks like a 90-degree angle so...

90 - 30 = 60

60 - 20 = 40

40 would be your final answer.

8 0

3 years ago

Read 2 more answers

Can someone PLEASE help me with this question?

disa [49]

9514 1404 393

Answer:

Step-by-step explanation:

The future value of the account is given by the formula ...

A = P(1 +r/12)^(12t) . . . . principal P invested at rate r for t years

Solving for t, we find ...

A/P = (1 +r/12)^(12t) . . . . . . . . . . . divide by P

log(A/P) = 12t·log(1 +r/12) . . . . . . take logs

Divide by the coefficient of t, then fill in the numbers.

t = log(A/P)/(12·log(1 +r/12)) = log(202800/93000)/(12·log(1 +.068/12))

t ≈ 11.497

It will take about 11 years for the account balance to reach the desired amount.

4 0

2 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$