Find an equation for the nth term of the arithmetic sequence.

2 answers:

8 0

1) To find out common equation for this sequence, it must be used n=1;2;3;4... etc.
2) to count a₁ use n=1. It is correct only for row $a_n=-3+-2(n-1)$ result is (-3)
3) if to verify others value for n:

Liono4ka [1.6K]3 years ago

7 0

An = -3 + -2 (n) is the answer. To verify this, start replacing the"n" with numbers like so: 0, 1, 2, 3, 4, 5....etc

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Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.

Darina [25.2K]

Answer:

$\cos( \theta) = \frac{5}{8} \\ {8}^{2} = {5}^{2} + {opp}^{2} \\ {opp}^{2} = 64 - 25 = 39 \\ opp = \sqrt{39} \\ \sin(\theta) = \frac{\sqrt{39}}{8} \\ \tan(\theta) = \frac{ \sqrt{39} }{5} \\ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{ \sqrt{39} }{5}} \\ \cot(\theta) = \frac{5}{ \sqrt{39} } \\ \csc( \theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{\sqrt{39}}{8}} \\ \csc( \theta) = \frac{8}{ \sqrt{39} } \\ \sec( \theta) = \frac{1}{\cos( \theta) } = \frac{1}{ \frac{5}{8} } \\ \sec( \theta) = \frac{8}{5}$

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2 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .