Answer:
Option 1
Step-by-step explanation:
Given expression representing the partial sum of the geometric series,
![\sum_{n=1}^{n=4}(125)(\frac{1}{5})^{n-1}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Bn%3D4%7D%28125%29%28%5Cfrac%7B1%7D%7B5%7D%29%5E%7Bn-1%7D)
Expression that represents the sum of a geometric series is,
![\sum_{n=1}^{n}(a)(r)^{n-1}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Bn%7D%28a%29%28r%29%5E%7Bn-1%7D)
Here, n = number of terms
a = first term
r = common ratio
By comparing both the expressions,
n = 4
a = 125
r = ![\frac{1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D)
From the given options,
Option 1
First term 'a' = 125
Common ratio 'r' =
= ![\frac{1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D)
Number of terms 'n' = 4
Therefore, Option 1 will be the correct option.
Answer:
35.770
Step-by-step explanation:
If the number behind the hundreths place is bigger than the hundreths place then you round up 1
Answer:
52.9°
Step-by-step explanation:
step 1
we know that
The total number of successful United States space launches that happened in the years 1957-1995 is equal to
![15+470+258+153+146=1,042](https://tex.z-dn.net/?f=15%2B470%2B258%2B153%2B146%3D1%2C042)
step 2
The numbers of successful United States space launches that happened in the years 1980-1989 is equal to
![153](https://tex.z-dn.net/?f=153)
step 3
The percentage of successful United States space launches that happened in the years 1980-1989 is equal to
![\frac{153}{1,042}(100)=14.68\%](https://tex.z-dn.net/?f=%5Cfrac%7B153%7D%7B1%2C042%7D%28100%29%3D14.68%5C%25)
step 4
Find out the measure of the central angle you would draw in a circle graph to represent the number of successful United States space launches that happened in the years 1980-1989
we know that
In a circle, 360 degrees represent the 100%
so
using proportion
Find out the measure of the central angle by a percentage of 14.68%
![\frac{360^o}{100\%}=\frac{x}{14.68\%}\\\\x=360^o(14.68\%)/100\%\\\\x= 52.9^o](https://tex.z-dn.net/?f=%5Cfrac%7B360%5Eo%7D%7B100%5C%25%7D%3D%5Cfrac%7Bx%7D%7B14.68%5C%25%7D%5C%5C%5C%5Cx%3D360%5Eo%2814.68%5C%25%29%2F100%5C%25%5C%5C%5C%5Cx%3D%2052.9%5Eo)
Answer:
and growth rate is 17%.
Step-by-step explanation:
The general exponential growth function is
...... (1)
where, a is initial value, r is growth rate and x is time.
The given function is
![f(x)=2(3)^{x}](https://tex.z-dn.net/?f=f%28x%29%3D2%283%29%5E%7Bx%7D)
Rose wants to manipulate the formula to an equivalent form that calculates seven times a year, not just once a year.
Using the properties of exponents the given function can be written as
![f(x)=2(3^{\frac{1}{7}})^{7x}](https://tex.z-dn.net/?f=f%28x%29%3D2%283%5E%7B%5Cfrac%7B1%7D%7B7%7D%7D%29%5E%7B7x%7D)
![f(x)=2(1.17)^{7x}](https://tex.z-dn.net/?f=f%28x%29%3D2%281.17%29%5E%7B7x%7D)
It can be rewritten as
.... (2)
On comparing (1) and (2) we get
r=0.17
Therefore, the required function is
and growth rate is 17%.
Luke is not correct
You need to do the problem in order of PEMDAS, so start with parentheses
2 x 2 = 4 + 6 = 10
Distribute the 2 on the outside
2 x 10 = 20
On the other side of the equation start with multiplication
2 x 2 = 4 + 1 = 5
5 ≠ 10
Luke is not correct