17.2/43= 0.4
Since the scale is 1:43, you would just divide the actual length by 43.
0.4:17.2 is congruent to 1:43.
43/17.2= 2.5
1/.04= 2.5
I hope this helps!
~kaikers
Answer:
Below.
Step-by-step explanation:
IQR is the same
Number of data points is the same.
Mode - can't tell
Range - different
First quartile - same
Median - different.
C the third one
Step by step
Answer:
A) ![\dfrac{11}{23}](https://tex.z-dn.net/?f=%5Cdfrac%7B11%7D%7B23%7D)
B) ![\dfrac{88}{345}](https://tex.z-dn.net/?f=%5Cdfrac%7B88%7D%7B345%7D)
Step-by-step explanation:
A survey of 46 college athletes found that
- 24 played volleyball,
- 22 played basketball.
A) If we pick one athlete survey participant at random, the probability they play basketball is
![P_1=\dfrac{22}{46}=\dfrac{11}{23}](https://tex.z-dn.net/?f=P_1%3D%5Cdfrac%7B22%7D%7B46%7D%3D%5Cdfrac%7B11%7D%7B23%7D)
B) If we pick 2 athletes at random (without replacement),
- the probability we get one volleyball player is
![\dfrac{24}{46}=\dfrac{12}{23};](https://tex.z-dn.net/?f=%5Cdfrac%7B24%7D%7B46%7D%3D%5Cdfrac%7B12%7D%7B23%7D%3B)
- the probability we get another basketball player is
(only 45 athletes left).
Thus, the probability we get one volleyball player and one basketball player is
![P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}](https://tex.z-dn.net/?f=P_2%3D%5Cdfrac%7B12%7D%7B23%7D%5Ccdot%20%5Cdfrac%7B22%7D%7B45%7D%3D%5Cdfrac%7B88%7D%7B345%7D)
Answer:
The correct answer is b.
Step-by-step explanation:
![16 {x}^{2} = - 16](https://tex.z-dn.net/?f=16%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20-%2016)
![{x}^{2} = - 1](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%3D%20%20-%201)
No real solutions.