Answer:
5 out of 7.
Step-by-step explanation:
There are seven days in one week. They are asking what is the probability of the person chosen not being born in a Tuesday or Wednesday. There are five other days, so there are five other options.
I hope I helped you!
Answer:
Median
mean>median
Step-by-step explanation:
When the data is skewed to right the suitable average is median. Median is suitable because it is less effected by extreme values and thus locate the center of the distribution perfectly. Here the salaries of basket players are skewed to right and the best measure of central tendency to measure the center of distribution is median.
When the frequency distribution is rightly skewed then the relationship of mean and median is that mean is greater than median that is Mean>median.
Hence when the distribution is skewed to right the best choice to measure the center of distribution is median and when the data is skewed to right mean is greater than median.
Answer:
![23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=23%5Csqrt%7B3%7D%5C%20un%5E2)
Step-by-step explanation:
Connect points I and K, K and M, M and I.
1. Find the area of triangles IJK, KLM and MNI:
![A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20IJ%5Ccdot%20JK%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%202%5Ccdot%203%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20KL%5Ccdot%20LM%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%208%5Ccdot%202%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20MN%5Ccdot%20NI%5Ccdot%20%5Csin%20120%5E%7B%5Ccirc%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%203%5Ccdot%208%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2%5C%5C%20%5C%5C%20%5C%5C)
2. Note that
![A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7B%5Ctriangle%20IJK%7D%3DA_%7B%5Ctriangle%20IAK%7D%3D%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20KLM%7D%3DA_%7B%5Ctriangle%20KAM%7D%3D4%5Csqrt%7B3%7D%5C%20un%5E2%20%5C%5C%20%5C%5C%20%5C%5CA_%7B%5Ctriangle%20MNI%7D%3DA_%7B%5Ctriangle%20MAI%7D%3D6%5Csqrt%7B3%7D%5C%20un%5E2)
3. The area of hexagon IJKLMN is the sum of the area of all triangles:
![A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2](https://tex.z-dn.net/?f=A_%7BIJKLMN%7D%3D2%5Ccdot%20%5Cleft%28%5Cdfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%2B4%5Csqrt%7B3%7D%2B6%5Csqrt%7B3%7D%5Cright%29%3D23%5Csqrt%7B3%7D%5C%20un%5E2)
Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.
Answer:
87 is the prime number
Step-by-step explanation:
the only factor of 87 is 1 and 87 which makes it prime
BRAINIEST PLZZZZZZ.