10, 30, 90 continue the pattern

2 answers:

7 0

The answer is 270, the pattern is multiplying by 3. So, 10 • 3 = 30 30 • 3 = 90 90 • 3 = 270, and so on. Hope this helps! Let me know if you need any further assistant.

Step2247 [10]3 years ago

3 0

Answer:

geometric sequence: 10, 30, 90, 270, 810, 2430, ...
quadratic sequence: 10, 30, 90, 190, 330, 510, ...

Step-by-step explanation:

Three terms of a sequence can be the first three terms of many different kinds of sequences. Here the ratios of terms are constant:

30/10 = 90/30 = 3

so the pattern could be that of a geometric sequence. In that sequence, each term is 3 times the previous one.

This pattern continues ...

10, 30, 90, 270, 810, 2430, ...

___

Another way to look at sequences of numbers is to consider their differences. Here the differences are not constant, and there aren't enough terms to tell how the differences change.

30 -10 = 20

90 -30 = 60

The second of these differences can be viewed several ways. One is to say that it is 40 more than the first of the differences. If each successive difference is 40 more than the last, then the pattern is described by a quadratic function:

$a_n=10+20(n-1)^2$

This pattern continues ...

10, 30, 90, 190, 330, 510, ...

You might be interested in

The Transamerica Pyramid in San Francisco is shaped like a square pyramid. It has a slant height of 856.1 feet and each side of

SpyIntel [72]

The Lateral Area Is 282513ft2

8 0

3 years ago

Determine mu Subscript x overbar and sigma Subscript x overbar from the given parameters of the population and sample size. mu e

Mademuasel [1]

Answer:

$\bar x \sim N(\mu_{\bar x}=86, \sigma_{\bar x}=3)$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable who represents the variable of interest. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =86,\sigma =24)$