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4 years ago

Write the rule for the nth term of the sequence. Then find a11-3.5, 3, 2.5, 2, 1.5, ...

Mathematics

1 answer:

RideAnS [48]4 years ago

6 0

Answer:

a₁₁ = -3/2

Step-by-step explanation:

3.5; 3 ; 2.5; 2 ; 1.5..............aₙ

the nth term

aₙ = -n/2 + 4

a₁₁ = (-0.5)(11) + 4

a₁₁ = -3/2

You might be interested in

Diego started an exercise program in which each week he works out at the gym between 4.5 and 6 hours and walks between 9 and 12

choli [55]

Answer:

$4.5$

Step-by-step explanation:

Let $x$ be time that Diego spends in gym and $y$ be walking distance.

Gym Time:

Minimum time of gym $=4.5\ hours$

Maximum time of gym $=6\ hours$

So the time $x$ will be greater than $4.5$ and less than $6$ .

$4.5$

Walking Distance:

Minimum Distance $=9\ miles$

Maximum Distance $=12\ miles$

So the walking distance will be greater than $9$ and less than $12$

$9$

Inequalities:

$4.5$

4 0

3 years ago

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

blsea [12.9K]

Answer:

$38.898 \leq \sigma^2 \leq 2792.356$

Now we just take square root on both sides of the interval and we got:

$6.237 \leq \sigma \leq 52.843$

Step-by-step explanation:

Data given and notation

34,59,61,71,59

We can calculate the sample standard deviation with the following formula:

$s^2= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$s= \sqrt{s^2}$

s=12.021 represent the sample standard deviation

$\bar x$ represent the sample mean

n=5 the sample size

Confidence=99% or 0.99

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical values.