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Over [174]
2 years ago
8

Hi, I need help with this math question. Can you pls help me?

Mathematics
2 answers:
frozen [14]2 years ago
3 0

The third term is 245 and the first term is 5 -> To get from the first term to the third term, we multiply it by 49.

Since in a geometric sequence, the difference between each term has to be a constant multiplier or a sequence of multipliers -> We can find the second number to be 5 x 7 = 35

So, this geometric sequence is 5 ; 35 ; 245 and so on.

-> The common ratio of this sequence is 7. (and sorry if the things above don't make sense)

podryga [215]2 years ago
3 0

Answer:

7

Step-by-step explanation:

Given:

  • First term = 5
  • Third term = 245

Let the second term be "x".

⇒ 5, x, 245...

Thus, the following equations:

  • 5 × (something) = x
  • x × (something) = 245

To find the common ratio between the three terms, we need to find the value of "x". Simplify both the equations to obtain the values of (something). Then compare both the values.

⇒ [5 × (something)] = x

⇒ [5 × (something)]/5 = x/5

⇒ (something) = x/5

⇒ x × (something) = 245

⇒ [x × (something)]/x = 245/x

⇒ (something) = 245/x

Now, let's compare both the results of (something).

⇒ x/5 = 245/x

⇒ x² = 5 x 245

⇒ x² = 1225

⇒ x = √1225 = 35

Substitute the second term into the sequence.

⇒ 5, x, 245 = 5, 35, 245

Looking at the sequence formed, we can tell that 7 is being multiplied to each term. Thus, 7 is the common ratio.

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4 years ago
A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their
Sever21 [200]

Answer:

A. Normal

B. Between 40.08 minutes and 43.92 minutes.

C. About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

Step-by-step explanation:

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

x% confidence interval:

A confidence interval is built from a sample, has bounds a and b, and has a confidence level of x%. It means that we are x% confident that the population mean is between a and b.

Question A:

By the Central Limit Theorem, a normal distribution.

Question B:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{12}{\sqrt{150}} = 1.92

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.92 = 40.08 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.92 = 43.92 minutes

Between 40.08 minutes and 43.92 minutes.

Question C:

x% confidence interval -> x% will contain the true population mean, (100-x)% wont.

So, 95% confidence interval:

About 95 percent of these confidence intervals will contain the true population mean number of minutes of concentration and about 5 percent will not contain the true population mean number of minutes of concentration.

3 0
3 years ago
Help plz it would mean alot to me
puteri [66]
I think the answer is a just from looking.
could you put a picture with the ruler next to all three sides please
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