Answer:
0.13774 inches ( approx ).
Step-by-step explanation:
Given,
The age of earth = 4.6 billion years = 4600000000 years,
The represented age of earth = 63,360 inches,
That is, scale factor in representation of age
![=\frac{\text{Represented age}}{\text{Actual age}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Ctext%7BRepresented%20age%7D%7D%7B%5Ctext%7BActual%20age%7D%7D)
![=\frac{ 63,360}{4600000000}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%2063%2C360%7D%7B4600000000%7D)
Now, the age of human civilization = 10000 years,
Thus, the represented age of human civilization = actual age × scale factor
![=10000\times \frac{ 63,360}{4600000000}](https://tex.z-dn.net/?f=%3D10000%5Ctimes%20%5Cfrac%7B%2063%2C360%7D%7B4600000000%7D)
![=\frac{633600000}{4600000000}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B633600000%7D%7B4600000000%7D)
![=0.137739130435](https://tex.z-dn.net/?f=%3D0.137739130435)
![\approx 0.13774\text{ inches}](https://tex.z-dn.net/?f=%5Capprox%200.13774%5Ctext%7B%20inches%7D)
First, solve for slope:
-7 - -8 = 1
-3 - -6 = 3
1/3 = m
Next, plug in one of your points into
y = (1/3) x + b
-7 = -1 + b
-6 = b
Answer:
y = (1/3)x-6
<h2>
Answer:</h2>
<em><u>Recursive equation for the pattern followed is given by,</u></em>
![a_{n}=a_{n-1}+(n-1)^{2}](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D)
<h2>
Step-by-step explanation:</h2>
In the question,
The number of interaction for 1 child = 0
Number of interactions for 2 children = 1
Number of interactions for 3 children = 5
Number of interaction for 4 children = 14
So,
We need to find out the pattern for the recursive equation for the given conditions.
So,
We see that,
![a_{1}=0\\a_{2}=1\\a_{3}=5\\a_{4}=14\\](https://tex.z-dn.net/?f=a_%7B1%7D%3D0%5C%5Ca_%7B2%7D%3D1%5C%5Ca_%7B3%7D%3D5%5C%5Ca_%7B4%7D%3D14%5C%5C)
Therefore, on checking, we observe that,
![a_{n}=a_{n-1}+(n-1)^{2}](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D)
On checking the equation at the given values of 'n' of, 1, 2, 3 and 4.
<u>At, </u>
<u>n = 1</u>
![a_{n}=a_{n-1}+(n-1)^{2}\\a_{1}=a_{1-1}+(1-1)^{2}\\a_{1}=0+0=0\\a_{1}=0](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D%5C%5Ca_%7B1%7D%3Da_%7B1-1%7D%2B%281-1%29%5E%7B2%7D%5C%5Ca_%7B1%7D%3D0%2B0%3D0%5C%5Ca_%7B1%7D%3D0)
which is true.
<u>At, </u>
<u>n = 2</u>
![a_{n}=a_{n-1}+(n-1)^{2}\\a_{2}=a_{2-1}+(2-1)^{2}\\a_{2}=a_{1}+1\\a_{2}=1](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D%5C%5Ca_%7B2%7D%3Da_%7B2-1%7D%2B%282-1%29%5E%7B2%7D%5C%5Ca_%7B2%7D%3Da_%7B1%7D%2B1%5C%5Ca_%7B2%7D%3D1)
Which is also true.
<u>At, </u>
<u>n = 3</u>
![a_{n}=a_{n-1}+(n-1)^{2}\\a_{3}=a_{3-1}+(3-1)^{2}\\a_{3}=a_{2}+4\\a_{3}=5](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D%5C%5Ca_%7B3%7D%3Da_%7B3-1%7D%2B%283-1%29%5E%7B2%7D%5C%5Ca_%7B3%7D%3Da_%7B2%7D%2B4%5C%5Ca_%7B3%7D%3D5)
Which is true.
<u>At, </u>
<u>n = 4</u>
![a_{n}=a_{n-1}+(n-1)^{2}\\a_{4}=a_{4-1}+(4-1)^{2}\\a_{4}=a_{3}+9\\a_{4}=14](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D%5C%5Ca_%7B4%7D%3Da_%7B4-1%7D%2B%284-1%29%5E%7B2%7D%5C%5Ca_%7B4%7D%3Da_%7B3%7D%2B9%5C%5Ca_%7B4%7D%3D14)
This is also true at the given value of 'n'.
<em><u>Therefore, the recursive equation for the pattern followed is given by,</u></em>
![a_{n}=a_{n-1}+(n-1)^{2}](https://tex.z-dn.net/?f=a_%7Bn%7D%3Da_%7Bn-1%7D%2B%28n-1%29%5E%7B2%7D)
Answer:
the client could expect a maximum loss of -0.054/year (-5,4%/year)
Step-by-step explanation:
since the 68-95-99.7 rule states that states probability that the anual return stays between 1 standard deviation from the mean is 68% , 2 standard deviations → 95% and 3 standard deviations → 99.7%
Then we are almost certain that the annual return will stay between 3 standard deviations from the mean.
Thus the most a client can loose is approximately at 3 standard deviations from the mean = 0.066 - 3*0.04 = -0.054 (-5,4%/year)