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An arithmetic sequence is defined by the general term tn = -5 + (n - 1)78, where n ∈N and n ≥ 1. What is the recursive formula o

Gnoma [55]

Answer:

C

Step-by-step explanation:

In general for arithmetic sequences, recursive formulas are of the form

aₙ = aₙ₋₁ + d,

and the explicit formula (like tₙ in your problem), are of the form

aₙ = a₁ + (n - 1)d,

where d is the common difference. So converting between the two of these isn't so bad. In this case, your problem wants you to have an idea of what t₁ is (well, every answer says it's -5, so there you are) and what tₙ₊₁ is. Using the formulas above and your given tₙ = -5 + (n - 1)78, we can see that the common difference is 78, so no matter what we get ourselves into, the constant being added on at the end should be 78. That automatically throws out answer choice D.

But to narrow it down between the rest of them, you want to use the general form for the recursive formula and substitute (n + 1) for every instance of n. This will let you find tₙ₊₁ to match the requirements of your answer choices. So

tₙ₊₁ = t₍ₙ₊₁₎₋₁ + d ... Simplify the subscript

tₙ₊₁ = tₙ + d

Therefore, your formula for tₙ₊₁ = tₙ + 78, which is answer choice C.

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3 years ago

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.