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ElenaW [278]
3 years ago
10

Find the 14th term of the geometric sequence 2, 6, 18, ...

Mathematics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

54

Step-by-step explanation:

its a sequence. they are multiplying by 3 each time.

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The primary reason why scientific notation is important is that it allows us to convert very large or very small numbers into much more manageable sizes. When these numbers are in scientific notation, it is much easier to work with them
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If 22% of a number is 154, then the
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A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G
dezoksy [38]

Answer:

a) The mean of a sampling distribution of \\ \overline{x} is \\ \mu_{\overline{x}} = \mu = 20. The standard deviation is \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

b) The standard normal z-score corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

c) The standard normal z-score corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

d) The probability \\ P(\overline{x}.

e) The probability \\ P(\overline{x}>23) = 1 - P(Z.

f)  \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

Step-by-step explanation:

We are dealing here with the concept of <em>a sampling distribution</em>, that is, the distribution of the sample means \\ \overline{x}.

We know that for this kind of distribution we need, at least, that the sample size must be \\ n \geq 30 observations, to establish that:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the distribution of the sample means follows, approximately, a <em>normal distribution</em> with mean, \mu, and standard deviation (called <em>standard error</em>), \\ \frac{\sigma}{\sqrt{n}}.

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> with \\ \mu = 0 and \\ \sigma = 1.

\\ Z \sim N(0, 1)

The variable Z is

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of \\ \overline{x} is the population mean \\ \mu = 20 or \\ \mu_{\overline{x}} = \mu = 20.

The standard deviation is the population standard deviation \\ \sigma = 16 divided by the root square of n, that is, the number of observations of the sample. Thus, \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

Part b

We are dealing here with a <em>random sample</em>. The z-score for the sampling distribution of \\ \overline{x} is given by [1]. Then

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{-4}{\frac{16}{8}}

\\ Z = \frac{-4}{2}

\\ Z = -2

Then, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

Part c

We can follow the same procedure as before. Then

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{3}{\frac{16}{8}}

\\ Z = \frac{3}{2}

\\ Z = 1.5

As a result, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

Part d

Since we know from [1] that the random variable follows a <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> for the corresponding \\ \overline{x} already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is \\ P(Z.

Therefore, the probability \\ P(\overline{x}.

Part e

We can follow a similar way than the previous step.

\\ P(\overline{x} > 23) = P(Z > 1.5)

For \\ P(Z > 1.5) using the <em>cumulative standard normal table</em>, we can find this probability knowing that

\\ P(Z1.5) = 1

\\ P(Z>1.5) = 1 - P(Z

Thus

\\ P(Z>1.5) = 1 - 0.9332

\\ P(Z>1.5) = 0.0668

Therefore, the probability \\ P(\overline{x}>23) = 1 - P(Z.

Part f

This probability is \\ P(\overline{x} > 16) and \\ P(\overline{x} < 23).

For finding this, we need to subtract the cumulative probabilities for \\ P(\overline{x} < 16) and \\ P(\overline{x} < 23)

Using the previous <em>standardized values</em> for them, we have from <em>Part d</em>:

\\ P(\overline{x}

We know from <em>Part e</em> that

\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z

\\ P(\overline{x} < 23) = P(Z1.5)

\\ P(\overline{x} < 23) = P(Z

\\ P(\overline{x} < 23) = P(Z

Therefore, \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

5 0
3 years ago
Plz help me I need help I’m being timed
lakkis [162]

Answer:

for the first one it is 112. For the second it is 150.

Step-by-step explanation:

Hope I helped :)

4 0
4 years ago
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