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Elza [17]
3 years ago
10

Phil's average is 0.250. This is about 5/6 as much as Joe's average. What is Joe's average? (Write answer is decimal form - do n

ot round).
Mathematics
1 answer:
LenaWriter [7]3 years ago
5 0

Let Joes average = x

Phil’s average is 0.250 and is 5/6X ( 5/6 of Joe’s)

5/6x = 0.250

Solve for x by dividing both sides by 5/6. When dividing by a fraction, flip the fraction over and multiply;

X = 0.250 x 6/5 = (0.250 x 6)/5 = 1.5/5 = 0.30

Joes average is 0.3

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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
A pet survey found that the ratio of dogs to cats is 2/5. Write and solve a proportion that shows the number of dogs
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So,

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The other proportion:
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Multiply the fraction by the appropriate form of 1
\frac{2}{5} * \frac{28}{28} =  \frac{56}{140}

There are 56 dogs when the number of cats is 140.
7 0
3 years ago
Amy joins a meal delivery service. Each family meal is $15, and there is a monthly flat shipping fee of $20. If Amy orders eight
8090 [49]

Answer: 140$

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4 0
2 years ago
Values of a and b<br><br><br><br><br><br><br><br><br><br><br> Mm
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