Answer:
2
Step-by-step explanation:
theres two numbers so i believe 2
Answer:
Range: 0.07
Median: 0.15
Mode: 0.15 and 0.16 (This is <u>bimodal</u>. It has 2 modes.)
Step-by-step explanation:
When you have a line and dots, that means that that number is repeated by the number of dots. So, since, 0.11 has 3 dots, when you write out all the numbers, you will write 0.11 3 times. (In the picture attached below, I have put every single number in order so you can see what I mean.)
<u>The Range</u> is the biggest number minus the smallest: 0.18 - 0.11 = <u>0.07</u>
<u>The Median</u> is the <u>m</u>iddle number: 0.15 (Look at picture)
<u>The Mode</u> is the number that occurs <u>m</u>ost <u>o</u>ften: 0.15 & 0.16
Having 2 modes is possible. They both just have to be repeated the same amount of times.
Answer:
144
Step-by-step explanation:
(6(2))(4)
aka (6*6)*4=144
Answer:
![\large\boxed{\bold{Q1}\ -\dfrac{128}{9}}\\\boxed{\bold{Q2}\ 2^{25}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cbold%7BQ1%7D%5C%20-%5Cdfrac%7B128%7D%7B9%7D%7D%5C%5C%5Cboxed%7B%5Cbold%7BQ2%7D%5C%202%5E%7B25%7D%7D)
Step-by-step explanation:
![\bold{Q1}\\\\\dfrac{-2(3^2\cdot3^3)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=-\dfrac{2(3^{2+3})^2}{3^{12}\cdot2^{-6}}=-\dfrac{2(3^5)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=-\dfrac{2\cdot3^{5\cdot2}}{3^{12}\cdot2^{-6}}=-\dfrac{2\cdot3^{10}}{3^{12}\cdot2^{-6}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=-2^{1-(-6)}\cdot3^{10-12}=-2^{7}\cdot3^{-2}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=-2^7\cdot\dfrac{1}{3^2}=-\dfrac{2^7}{3^2}=-\dfrac{128}{9}](https://tex.z-dn.net/?f=%5Cbold%7BQ1%7D%5C%5C%5C%5C%5Cdfrac%7B-2%283%5E2%5Ccdot3%5E3%29%5E2%7D%7B3%5E%7B12%7D%5Ccdot2%5E%7B-6%7D%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5C%3D-%5Cdfrac%7B2%283%5E%7B2%2B3%7D%29%5E2%7D%7B3%5E%7B12%7D%5Ccdot2%5E%7B-6%7D%7D%3D-%5Cdfrac%7B2%283%5E5%29%5E2%7D%7B3%5E%7B12%7D%5Ccdot2%5E%7B-6%7D%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%3D-%5Cdfrac%7B2%5Ccdot3%5E%7B5%5Ccdot2%7D%7D%7B3%5E%7B12%7D%5Ccdot2%5E%7B-6%7D%7D%3D-%5Cdfrac%7B2%5Ccdot3%5E%7B10%7D%7D%7B3%5E%7B12%7D%5Ccdot2%5E%7B-6%7D%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%5Cdfrac%7Ba%5En%7D%7Ba%5Em%7D%3Da%5E%7Bn-m%7D%5C%5C%5C%5C%3D-2%5E%7B1-%28-6%29%7D%5Ccdot3%5E%7B10-12%7D%3D-2%5E%7B7%7D%5Ccdot3%5E%7B-2%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%7B-n%7D%3D%5Cdfrac%7B1%7D%7Ba%5En%7D%5C%5C%5C%5C%3D-2%5E7%5Ccdot%5Cdfrac%7B1%7D%7B3%5E2%7D%3D-%5Cdfrac%7B2%5E7%7D%7B3%5E2%7D%3D-%5Cdfrac%7B128%7D%7B9%7D)
![\bold{Q2}\\\\\dfrac{4^2}{4^{-7}}\cdot(2^4\cdot2^3)\cdot(3^{-2})^0\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m},\ a^n\cdot a^m=a^{n+m},\ a^0=1\\\\=4^{2-(-7)}\cdot2^{4+3}\cdot1=4^{9}\cdot2^7=(2^2)^{9}\cdot2^7\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{2\cdot9}\cdot2^7=2^{18}\cdot2^7\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{18+7}=2^{25}](https://tex.z-dn.net/?f=%5Cbold%7BQ2%7D%5C%5C%5C%5C%5Cdfrac%7B4%5E2%7D%7B4%5E%7B-7%7D%7D%5Ccdot%282%5E4%5Ccdot2%5E3%29%5Ccdot%283%5E%7B-2%7D%29%5E0%5Cqquad%5Ctext%7Buse%7D%5C%20%5Cdfrac%7Ba%5En%7D%7Ba%5Em%7D%3Da%5E%7Bn-m%7D%2C%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%2C%5C%20a%5E0%3D1%5C%5C%5C%5C%3D4%5E%7B2-%28-7%29%7D%5Ccdot2%5E%7B4%2B3%7D%5Ccdot1%3D4%5E%7B9%7D%5Ccdot2%5E7%3D%282%5E2%29%5E%7B9%7D%5Ccdot2%5E7%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%3D2%5E%7B2%5Ccdot9%7D%5Ccdot2%5E7%3D2%5E%7B18%7D%5Ccdot2%5E7%5Cqquad%5Ctext%7Buse%7D%5C%20a%5En%5Ccdot%20a%5Em%3Da%5E%7Bn%2Bm%7D%5C%5C%5C%5C%3D2%5E%7B18%2B7%7D%3D2%5E%7B25%7D)
Ok Gudrunverizuensievr sorry