If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
Find the interquartile range for the data {5, 7, 9, 5, 6, 6, 6, 11, 11, 3, 3}
RoseWind [281]
Answer:
4
Step-by-step explanation:
i dont really know how to explain i used an algebra calculator
10. Find which one of the given fractions are value closer to 1.
First = 7/8
Second = 11/10
Let’s start solving
Let us start with the first value which is 7/8
=> 7 needs to be divided with 8
=> 7 / 8 = 0.875
Now, let’s solve the second value which is 11/10
=> same with first solution., divide 11 by 10
=> 11 / 10
=> 1.1
Thus, 11/10 is the has the nearest value to 1.
The answer is 5-3/2x. Hope this helps :)
Since they meet at the $105 mark the answer would be $105
