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nata0808 [166]
2 years ago
4

Keith spends 1/6 of his savings on a magazine 2/5 of the remainder on a storybook what fraction of his savings is left

Mathematics
2 answers:
m_a_m_a [10]2 years ago
5 0
6/6 - 1/6 = 5/6

5/6 - 2/5
= ( 25 - 12)/30

= 13/30
ddd [48]2 years ago
3 0


 he spent 2/5*25=10
So he has spent 5+10=15
15/30=1/2
so he has 1/2 or $15 dollars left.

I hope it correct



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Suppose the area of a circle is 254.4696 square inches. Whats the radius of the circle? Use pie=3.1416. A. 18 in B. 81 in. C. 9
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Answer:

radius = 9 inches

Step-by-step explanation:

Area = \pi r^{2}

r = radius

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so we have:

r^{2} = \frac{A}{\pi} \\\\r = \sqrt{\frac{A}{\pi}} \\\\r = \sqrt{\frac{254.4696}{3.1416}} \\\\r = \sqrt{81}

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2 years ago
Solve the trigonometric equation in the interval [0, 2π). Give the exact value, if possible; otherwise, round your answer to two
svp [43]

Answer:

θ∈{\frac{\pi }{8},\frac{5\pi }{8},\frac{9\pi }{8},\frac{13\pi }{8}}

Explanation:

The given equation is

sin(2\theta )-cos(2\theta )=0

\Rightarrow sin(2\theta )=cos(2\theta )\\\\\therefore \frac{sin(2\theta )}{cos(2\theta )}=1\\\\tan(2\theta )=1\\\\\therefore 2\theta =n\pi +\frac{\pi}{4}\\\\\therefore \theta =\frac{n\pi }{2}+\frac{\pi }{8}

Applying values on 'n' we obtain values of θ that beling to [0,2π)

For n=0, θ=\frac{\pi }{8}

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For n=2,θ =\frac{9\pi }{8}

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2 years ago
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2 years ago
The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second. Use the central limi
Mila [183]

Answer:

0.048 is the probability that more than 950 message arrive in one minute.

Step-by-step explanation:

We are given the following information in the question:

The number of messages arriving at a multiplexer is a Poisson random variable with mean 15 messages/second.

Let X be the number of messages arriving at a multiplexer.

Mean = 15

For poison distribution,

Mean = Variance = 15

\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{15} = 3.872

From central limit theorem, we have:

z = \displaystyle\frac{x-n\mu}{\sigma\sqrt{n}}

where n is the sample size.

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P(x > 950)

P( x > 950) = P( z > \displaystyle\frac{950 - (60)(15)}{\sqrt{(15)(60)}}) = P(z > 1.667)

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Calculation the value from standard normal z table, we have,  

P(x > 950) = 1 - 0.952 = 0.048

0.048 is the probability that more than 950 message arrive in one minute.

3 0
2 years ago
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