Answer:
A) x degrees + 65 degrees = 180 degrees
B) x degrees = z degrees
C) y degrees = 65 degrees
Step-by-step explanation:
Refer the attached figure
So,![\angle AOB = \angle COD (\text{Vertically opposite angles})](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%20%5Cangle%20COD%20%20%28%5Ctext%7BVertically%20opposite%20angles%7D%29)
![\angle AOB = 65^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%2065%5E%7B%5Ccirc%7D)
![\angle COD = y](https://tex.z-dn.net/?f=%5Cangle%20COD%20%3D%20%20y)
So, ![y=65^{\circ}](https://tex.z-dn.net/?f=y%3D65%5E%7B%5Ccirc%7D)
![\angle AOB + \angle BOC = 180^{\circ} (\text{Linear pair})](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%2B%20%5Cangle%20BOC%20%3D%20180%5E%7B%5Ccirc%7D%20%28%5Ctext%7BLinear%20pair%7D%29)
![\angle AOB = 65^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20AOB%20%3D%2065%5E%7B%5Ccirc%7D)
![\angle BOC = x](https://tex.z-dn.net/?f=%5Cangle%20BOC%20%3D%20x)
So, ![x+65^{\circ}=180^{\circ}](https://tex.z-dn.net/?f=x%2B65%5E%7B%5Ccirc%7D%3D180%5E%7B%5Ccirc%7D)
![\angle BOC = \angle AOD (\text{Vertically opposite angles})](https://tex.z-dn.net/?f=%5Cangle%20BOC%20%3D%20%5Cangle%20AOD%20%28%5Ctext%7BVertically%20opposite%20angles%7D%29)
So, x = z
So, Option A ,B and C are true for the values of x, y, and z
Answer:
if AB II CD then AGE should be also 60
Answer:
C = 11.75 + 0.20*x
Step-by-step explanation:
C = 11.75 + 0.20*x
Answer:
15
Step-by-step explanation:
We know that heart rate measurements are normally distributed with mean μ=110. We have to find standard deviation whereas minimum and maximum value of data are 65 and 155 beats per min.
We know that by the empirical rule, 99.7% of value lies within 3 standard deviation from the mean. We can see that 99.7% covers approximately all the data and we can assume that all the data lies within 3 standard deviation. Now we check for each value whether the interval μ±3*σ contains minimum value 65 and maximum value 155 beats per min or not.
A. 5
μ±3*σ
110±3*5
110±15
(95,125)
B. 15
μ±3*σ
110±3*15
110±45
(65,155)
We can see that for standard deviation=15 , μ±3*σ contains minimum value=65 and maximum value=155.
C. 90
μ±3*σ
110±3*90
110±270
(-160,380)
Thus, the standard deviation of the distribution is most likely to be 15.