Which best describes the relationship between the successive terms in the sequence below? –3.2, 4.8, –7.2, 10.8, …

2 answers:

6 0

Since you are not given the following choices, by analyzing this series, it is a geometric progression because their common ratio is constant. Proofing that
4.8/(-3.2) = -1.5, also
-7.2/(4.8) = -1.5, and also
10.8/(-7.2) = -1.5

taurus [48]3 years ago

5 0

Solution:

we have been asked to find the relationship between the successive terms in the sequence below? –3.2, 4.8, –7.2, 10.8, …

As we can see the given successive terms in the sequence is following a particular pattern as shown below:

$\frac{4.8}{-3.2}=\frac{-7.2}{4.8}=\frac{10.8}{-7.2}=-1.5$

Its following properties of geometric sequence. Because the ratios of the successive terms is constant. So its a geometric progression.

Hence the given successive terms in the sequence is a geometric sequence.

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The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$