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denis23 [38]
3 years ago
12

Find the sum of the geometric sequence. 1, one divided by two, one divided by four, one divided by eight, one divided by sixteen

Mathematics
2 answers:
krek1111 [17]3 years ago
8 0

The equation for a finite geometric sequence is Sn=a1((1-r^n)/(1-r))

a1=1

r=.5/1=.5

n=5

Using the equation:

Sn=a1((1-r^n)/(1-r))=1((1-.5^5)/(1-.5))=1((1-.03125)/(1-.5))=1(.96875/.5)=1*1.9375=1.9375

Answer Choices:

A. 1/12 or .083

B. 93

C. -1/48 or -.02083

D. 31/16 or 1.9375

The Answer is D. 31/16

igomit [66]3 years ago
5 0
a_1=1;\ a_2=\dfrac{1}{2};\ a_3=\dfrac{1}{4};\ a_4=\dfrac{1}{8};\ a_5=\dfrac{1}{16};\ ...\\\\d=a_2:a_1\\\\d=\dfrac{1}{2}:1=\dfrac{1}{2}\\\\|d| \ \textless \  1\ therefore\ the\ sum\ S\ of\ the\ geometric\ sequence\ is\ equal:\\\\S=\dfrac{a_1}{1-q}\\\\subtitute\\\\S=\dfrac{1}{1-\frac{1}{2}}=\dfrac{1}{\frac{1}{2}}=1\cdot\dfrac{2}{1}=2\\\\\\Answer:\boxed{S=2}
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A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 33.9 wee
MArishka [77]

Answer:

P(35.3 < M < 35.4) = 0.0040.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 33.9, \sigma = 6.7, n = 119, s = \frac{6.7}{\sqrt{119}} = 0.6142

Find the probability that a single randomly selected value is between 35.3 and 35.4

This is the pvalue of Z when X = 35.4 subtracted by the pvalue of Z when X = 35.3. So

X = 35.4

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{35.4 - 33.9}{0.6142}

Z = 2.44

Z = 2.44 has a pvalue of 0.9927

X = 35.3

Z = \frac{X - \mu}{s}

Z = \frac{35.3 - 33.9}{0.6142}

Z = 2.28

Z = 2.28 has a pvalue of 0.9887

0.9927 - 0.9887 = 0.0040

So the answer is:

P(35.3 < M < 35.4) = 0.0040.

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3 years ago
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Answer:

c=38 degrees

Step-by-step explanation:

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Width is 14 meters

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A spinner has five congruent sections, one each of blue, green, red, orange, and yellow. Yuri spins the spinner 10 times and rec
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