F(1) = 4 is given

f(2) = 2*f(2-1)+3
= 2*f(1)+3
= 2*4+3
= 8+3
= 11
f(3) = 2*f(3-1)+3
= 2*f(2)+3
= 2*11+3
= 22+3
= 25
f(4) = 2*f(4-1)+3
= 2*f(3)+3
= 2*25+3
= 50+3
= 53
f(5) = 2*f(5-1)+3
= 2*f(4)+3
= 2*53+3
= 106+3
= 109

f(1) = 4
f(2) = 11
f(3) = 25
f(4) = 53
f(5) = 109

8 0

3 years ago

What is the recursive formula for the geometric sequence ? 6,-24,96,-384,...

irinina [24]

Answer:

$the \: recursive \: formula \: is \\ tn = 6( {4}^{n - 1} )$

Answer is given below with explanations.

Step-by-step explanation:

$the \: given \: geometric \: sequence \: is \\ 6, \: - 24, \: 96, \: - 384 \\ the \: first \: term \: is \: 6 \: and \: the \: common \: ratio \: is \: - 4 \\ the \: {n}^{th} term \: of \: the \: geometric \: sequence \: is \\ tn = a {r}^{n - 1} \\ here \: first \: term \:( a) = 6 \\ common \: ratio \: (r) = - 4 \\ on \: substituting \: the \: values \: in \: formula \\ tn = 6( { - 4}^{n - 1} ) \\ the \: above \: mentioned \: formula \: is \: \\ the \: recursive \: formula \: \\ for \: geometric \: sequence.$

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5 0

3 years ago

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

8 0

3 years ago