Answer:
$6500
Step-by-step explanation:
A bank loaned out $18,000
Let x be the part of amount at the rate of 6 % per year
So, Remaining amount = 18000-x
So, the part of amount at the rate of 14 % per year = 18000-x
the interest received in one year totaled $2000
Formula : ![SI = \frac{P \times R \times T}{100}](https://tex.z-dn.net/?f=SI%20%3D%20%5Cfrac%7BP%20%5Ctimes%20R%20%5Ctimes%20T%7D%7B100%7D)
P is the principal
R = rate of interest
T = Time
SI at 6%+SI at 14% = 2000
![\frac{x \times 6 \times 1 }{100}+\frac{(18000-x) \times 14 \times 1 }{100}=2000](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%5Ctimes%206%20%5Ctimes%201%20%7D%7B100%7D%2B%5Cfrac%7B%2818000-x%29%20%5Ctimes%2014%20%5Ctimes%201%20%7D%7B100%7D%3D2000%20)
![\frac{-2 }{25}(x-31500)=2000](https://tex.z-dn.net/?f=%5Cfrac%7B-2%20%7D%7B25%7D%28x-31500%29%3D2000)
![x=6500](https://tex.z-dn.net/?f=x%3D6500%20)
Hence the bank loan out at 6% is $6500
i would say -6a^6*9
do 3 x -2 and get -6 then leave the exponent as 6 because you dont do anything with them but leave them the same
It would not be isosceles it would be scalene because a scalene doenst have the same sides all it sides are different
The value of n is a number less than -6
Given:
The data values are
11, 12, 10, 7, 9, 18
To find:
The median, lowest value, greatest value, lower quartile, upper quartile, interquartile range.
Solution:
We have,
11, 12, 10, 7, 9, 18
Arrange the data values in ascending order.
7, 9, 10, 11, 12, 18
Divide the data in two equal parts.
(7, 9, 10), (11, 12, 18)
Divide each parenthesis in 2 equal parts.
(7), 9, (10), (11), 12, (18)
Now,
Median = ![\dfrac{10+11}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B10%2B11%7D%7B2%7D)
=
=
Lowest value = 7
Greatest value = 18
Lower quartile = 9
Upper quartile = 12
Interquartile range (IQR) = Upper quartile - Lower quartile
= 12 - 9
= 3
Therefore, median is 10.5, lowest value is 7, greatest value is 18, lower quartile 9, upper quartile 12 and interquartile range is 3.