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IrinaK [193]
3 years ago
11

Given a geometric sequence in the table below, create the explicit formula and list any restrictions to the domain.

Mathematics
2 answers:
weeeeeb [17]3 years ago
4 0

Answer:

B

Step-by-step explanation:

given the first 3 terms of the geometric sequence - 4, 20, - 100

with r = \frac{-100}{20} = \frac{20}{-4} = - 5

the n th term formula ( explicit formula ) is

a_{n} = a_{1} r^{n-1}

here r = - 5 and a_{1} = - 4, hence

a_{n} = - 4 (-5)^{n-1} with n ≥ 1





pishuonlain [190]3 years ago
4 0

Answer:

answer would be b.

Step-by-step explanation:

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emmasim [6.3K]

The empirical rule you're referring to is the 68-95-99.7 rule, which asserts that for a normal (bell-shaped) distribution, approximately 68% of the distribution lies within 1 standard deviation of the mean; 95% lies within 2 standard deviations of the mean; and 99.7% lies within 3 standard deviations of the mean.

Let X be the random variable denoting vehicle speeds along this highway. We want to find P(61. To use the rule, we need to rephrase this probability in terms of the mean and standard deviation.

Notice that 61=70-9=70-3\cdot3, and 79=70+9=70+3\cdot3. In other words, 61 and 79 both lie exactly 3 standard deviations away from the mean, so P(61.

8 0
3 years ago
How to know if a function is periodic without graphing it ?
zhenek [66]
A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

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More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

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4 0
3 years ago
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Lapatulllka [165]

Answer:

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