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romanna [79]
3 years ago
11

Perform the operation 0.5x0.5+0.5/0.5=

Mathematics
2 answers:
natima [27]3 years ago
7 0
<span>0.5x0.5+0.5/0.5=0,25+1=1,25</span>
oksano4ka [1.4K]3 years ago
3 0
I believe .5x.5 is .25 and .5/.5 is 1 so add 1 and .25 to get 1.25. Hopefully? (:
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Find the value of sin( \frac{5 \pi }{3} )

Rewrite as sin( \frac{6 \pi - \pi }{3} )

sin( \frac{6}{3} \pi -  \frac{ \pi }{3}  )

sin(2 \pi - \frac{ \pi }{3} )

since sin(π/3) = √3/2

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-sin( \frac{ \pi }{3} )

- \frac{ \sqrt{3} }{2}
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3 years ago
What is the area, please help me!
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Answer:

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2 years ago
Read 2 more answers
A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat
zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 200, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5

a. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 200 + 5 = 205 subtracted by the pvalue of Z when X = 200 - 5 = 195.

Due to the Central Limit Theorem, Z is:

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

X = 205

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

Z = \frac{205 - 200}{5}

Z = 1

Z = 1 has a pvalue of 0.8413.

X = 195

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

Z = \frac{195 - 200}{5}

Z = -1

Z = -1 has a pvalue of 0.1587.

0.8413 - 0.1587 = 0.6426

0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

This is the pvalue of Z when X = 210 subtracted by the pvalue of Z when X = 190.

X = 210

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

Z = \frac{210 - 200}{5}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 195

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

Z = \frac{190 - 200}{5}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

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3 years ago
Two math students were asked to write an exponential growth equation that had a starting value of 300 and a growth rate of 2%. P
Gemiola [76]

Answer:

Pierre is right

Step-by-step explanation:

The correct formula for Exponential growth rate is given as:

y = a( 1 + r) ^t

Where

y = Amount after time t

a = Initial amount

r = Growth rate

t = time

From the question

a = 300

r = 2% = 0.02

Hence, our exponential growth rate =

y = 300( 1 + 0.02)^t

y = 300( 1.02)^t

Therefore, Pierre is right

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2 years ago
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Julli [10]
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