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garik1379 [7]
3 years ago
5

Sue solved an equation below. Check her work.

Mathematics
1 answer:
wolverine [178]3 years ago
4 0

Answer:

You cant see the equation

Step-by-step explanation:

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4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

6 0
3 years ago
Laura creates a rectangular prism with wooden cubes. Each cube has an edge length of 3/4 inch. she uses a total of 240 cubes. th
kirill115 [55]

Answer:00.24

explanation:

Just add the decimal form and multiply the 240 with the decimal of the 3/4

6 0
3 years ago
Please help !! If you help i’ll give 5 stars
sladkih [1.3K]

Answer:

Imaginary

Step-by-step explanation:

6 0
3 years ago
You will have 27 points if you can help me with this please!!!!
KatRina [158]

Answer:1: P 2: d 3:a

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Medical scientists study the effect of acute infection on tissue-specific immunity. In a collection of experiments under the sam
Serjik [45]

Answer:

The 95% confidence interval  for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

In a collection of experiments under the same conditions, 44 of 75 mice test positive for lymphadenopathy. This means that n = 75 and \pi = \frac{44}{75} = 0.586.

Compute a 95% confidence interval for the true proportion of mice that will test positive under similar conditions.

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 - 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.5291

The upper limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.586 + 1.96\sqrt{\frac{0.586*0.414}{75}} = 0.6429

The 95% confidence interval  for the true proportion of mice that will test positive under similar conditions is (0.5291, 0.6429).

7 0
3 years ago
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