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anyanavicka [17]
3 years ago
10

What is 3n+1 50th term?

Mathematics
1 answer:
Fofino [41]3 years ago
4 0
1<span>, 4, 7, 10 is a sequence starting with </span>1<span>. You get the next </span>term<span> by adding 3 to the previous </span>term. You are often asked to find a formula for the nth term<span>. As the common difference is 3, try </span>3n<span>. When n = </span>1<span>, </span>3n<span> = 3, and we subtract 2 to make the first </span>term<span> correct.</span>
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Show that 155 can be expressed as the sum of a power of 2<br> and a cube number.<br> I
Mice21 [21]

Answer:

155 = 2^7 + 3^3

Step-by-step explanation:

3 cubed = 27

2 to the 7th power = 128

27 plus 128 = 155

have a shiny day.

5 0
3 years ago
Concerns about climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g., from rene
Veronika [31]

Answer:

99% of the sample means will fall between 0.93288 and 0.94112.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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.

The true mean is .9370 with a standard deviation of 0.0090

This means that \mu = 0.9370, \sigma = 0.0090

Sample of 32:

This means that n = 32, s = \frac{0.009}{32} = 0.0016

Within what interval will 99 percent of the sample means fall?

Between the 50 - (99/2) = 0.5th percentile and the 50 + (99/2) = 99.5th percentile.

0.5th percentile:

X when Z has a pvalue of 0.005. So X when Z = -2.575.

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

By the Central Limit Theorem

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

-2.575 = \frac{X - 0.9370}{0.0016}

X - 0.9370 = -2.575*0.0016

X = 0.93288

99.5th percentile:

X when Z has a pvalue of 0.995. So X when Z = 2.575.

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

2.575 = \frac{X - 0.9370}{0.0016}

X - 0.9370 = 2.575*0.0016

X = 0.94112

99% of the sample means will fall between 0.93288 and 0.94112.

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3 years ago
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IceJOKER [234]

Answer:

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Step-by-step explanation:

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simplify

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1/6 may be but in decimal i got 0.6
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