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MissTica
3 years ago
10

The Super Bounce brand of bouncy balls rebounds to 85% of the height from which it was dropped. Write both the explicit and recu

rsive formulas h(n) that describes the height, h(n), of the bounce after n bounces given an initial height of 4ft.
Mathematics
2 answers:
jarptica [38.1K]3 years ago
8 0

Answer:

Explicit formula is h(n)=4(0.85)^{n-1}.

Recursive formula is h_n=0.85h_{n-1}

Step-by-step explanation:

Step 1

In this step we first find the explicit formula for the height of the ball.To find the explicit formula we use the fact that the bounces form a geometric sequence. A geometric sequence has the general formula ,a_{n+1}=ar^{n-1}. In this case the first term a_o=4, the common ratio r=0.85 since the ball bounces back to 0.85 of it's previous height.

We can write the explicit formula as,

h(n)=4(0.85)^{n-1}.

Step 2

In this step we find the recursive formula for the height of the ball after each bounce. Since the ball bounces to 0.85 percent of it's previous height, we know that to get the next term in the sequence, we have to multiply the previous term by the common ratio.  The general fomula for a geometric sequene is a_n=a_{n-1}\times r.

With the parameters given in this problem, we write the general term of the sequence as ,

h(1)=4\\h(n)=h_{n-1}\times 0.85.


PtichkaEL [24]3 years ago
6 0

Answer:

Recursive formula h_n = 0.85 \times h_{n-1}

Explicit formula h_n = 4 \times 0.85^n

Step-by-step explanation:

An explicit formula allows you to find the value of any term in the sequence. A recursive formula allows you to find the value of the nth term if you know the value of the (n-1)th term in the sequence.

<em>Recursive formula</em>

Given the initial height (h_0), the next height (h_1) is 85% of it, the following height is 85% of the new height, and so on. Mathematically:

h_n = 0.85 \times h_{n-1}

h_0 = 4 \; ft

Where (h_n) is the nth height, given the previous one (h_{n-1})

<em>Explicit formula</em>

initial height: 4 ft

n: number of bounces

h(n): height after n bounces

n    h(n)

1    4*0.85 = 3.4 ft

2    3.4*0.85 = 2.89 ft

3    2.89*0.85 = 2.4565 ft

...  ...

n    4*0.85^n

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Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

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The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

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s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

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S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

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It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

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   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

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*Use a <em>t</em>-table for the values.

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