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rodikova [14]
3 years ago
8

Make a conjecture about the next item in the sequence. 6, 9, 7, 10, 8

Mathematics
1 answer:
ozzi3 years ago
3 0

Answer:

6,9,7,10,8,11..

Step-by-step explanation:

If you notice that in the given sequence 6, 9, 7, 10, 8 the first element is 6 and then 3 is added in 6 to make it 9. Then 2 is subtracted from 9 which makes it 7 and then again 3 is added in 7 which makes it 10. After that again 2 is subtracted from 10 to make it 8. Now by following this method now we will add 3 in 8 which will give us the next item which is 11.

Thus the sequence we get is:

6,9,7,10,8,11...

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The answer is 11x-11
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Please put your answer in simplest form<br><br> 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = ?
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The current height of the bird is -16/3

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1 year ago
A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor dec
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Answer:

(a) The usual load is not 13 credits.

(b) The probability that a a student at this college takes 16 or more credits is 0.1093.

Step-by-step explanation:

According to the Central limit theorem, if a large sample (<em>n</em> ≥ 30) is selected from an unknown population then the sampling distribution of sample mean follows a Normal distribution.

The information provided is:

Min.=8\\Q_{1}=13\\Median=14\\Mean=13.65\\SD=1.91\\Q_{3}=15\\Max.=18

The sample size is, <em>n</em> = 100.

The sample size is large enough for estimating the population mean from the sample mean and the population standard deviation from the sample standard deviation.

So,

\mu_{\bar x}=\bar x=13.65\\SE=\frac{s}{\sqrt{n}}=\frac{1.91}{\sqrt{100}}=0.191

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Construct a (1 - <em>α</em>) % confidence interval for population mean to check the claim.

The (1 - <em>α</em>) % confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\times SE

For 5% level of significance the two tailed critical value of <em>z</em> is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct the 95% confidence interval as follows:

CI=\bar x\pm z_{\alpha/2}\times SE\\=13.65\pm (1.96\times0.191)\\=13.65\pm0.3744\\=(13.2756, 14.0244)\\=(13.28, 14.02)

As the null value, <em>μ</em> = 13 is not included in the 95% confidence interval the null hypothesis will be rejected.

Thus, it can be concluded that the usual load is not 13 credits.

(b)

Compute the probability that a a student at this college takes 16 or more credits as follows:

P(X\geq 16)=P(\frac{X-\mu}{\sigma}\geq \frac{16-13.65}{1.91})\\=P(Z>1.23)\\=1-P(Z

Thus, the probability that a a student at this college takes 16 or more credits is 0.1093.

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