The true statement about the circle with center P is that triangles QRP and STP are congruent, and the length of the minor arc is 11/20π
<h3>The circle with center P</h3>
Given that the circle has a center P
It means that lengths PQ, PR, PS and PT
From the question, we understand that QR = ST.
This implies that triangles QRP and STP are congruent.
i.e. △QRP ≅ △STP is true
<h3>The length of the minor arc</h3>
The given parameters are:
Angle, Ф = 99
Radius, r = 1
The length of the arc is:
L = Ф/360 * 2πr
So, we have:
L = 99/360 * 2π * 1
Evaluate
L = 198/360π
Divide
L = 11/20π
Hence, the length of the minor arc is 11/20π
Read more about circle and arcs at:
brainly.com/question/3652658
#SPJ1
3z-4=6z-17
-3z-4=-17 (subtracted the 6z from both sides)
-3z=-13 (added the 4 to both sides)
z=
![\frac{-13}{-3z}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-13%7D%7B-3z%7D%20)
(divided both sides by -3z)
z=
13 is the answer for ur question
Answer:
![s = 0.0394](https://tex.z-dn.net/?f=s%20%3D%200.0394)
Step-by-step explanation:
Given:
0.612 0.523 0.606 0.631 0.584 0.592 0.644 0.597 0.639 0.607 0.564 0.673
Required
Calculate the sample standard deviation
First, calculate the mean
![\bar x = \frac{\sum x}{n}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%5Csum%20x%7D%7Bn%7D)
![\bar x = \frac{0.612 +0.523 +0.606 +0.631+ 0.584 + 0.592+ 0.644 +0.597 +0.639 +0.607 +0.564+ 0.673}{12}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B0.612%20%2B0.523%20%2B0.606%20%2B0.631%2B%200.584%20%2B%200.592%2B%200.644%20%20%2B0.597%20%2B0.639%20%20%2B0.607%20%2B0.564%2B%200.673%7D%7B12%7D)
![\bar x = \frac{7.272}{12}](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B7.272%7D%7B12%7D)
![\bar x = 0.606](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%200.606)
The sample standard deviation is then calculated using:
![s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum%28x%20-%20%5Cbar%20x%29%5E2%7D%7Bn-1%7D%7D)
![\sum (x - \bar x)^2 = (0.612 -0.606)^2+(0.523 -0.606)^2+(0.606 -0.606)^2+(0.631 -0.606)^2+(0.584 -0.606)^2+(0.592 -0.606)^2+(0.644 -0.606)^2+(0.597 -0.606)^2+(0.639 -0.606)^2+(0.607 -0.606)^2+(0.564 -0.606)^2+(0.673 -0.606)^2](https://tex.z-dn.net/?f=%5Csum%20%28x%20-%20%5Cbar%20x%29%5E2%20%3D%20%280.612%20-0.606%29%5E2%2B%280.523%20-0.606%29%5E2%2B%280.606%20-0.606%29%5E2%2B%280.631%20-0.606%29%5E2%2B%280.584%20-0.606%29%5E2%2B%280.592%20-0.606%29%5E2%2B%280.644%20-0.606%29%5E2%2B%280.597%20%20-0.606%29%5E2%2B%280.639%20-0.606%29%5E2%2B%280.607%20-0.606%29%5E2%2B%280.564%20-0.606%29%5E2%2B%280.673%20-0.606%29%5E2)
![\sum (x - \bar x)^2 = 0.017098](https://tex.z-dn.net/?f=%5Csum%20%28x%20-%20%5Cbar%20x%29%5E2%20%3D%200.017098)
So, we have:
![s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum%28x%20-%20%5Cbar%20x%29%5E2%7D%7Bn-1%7D%7D)
![s = \sqrt{\frac{0.017098}{12-1}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7B0.017098%7D%7B12-1%7D%7D)
![s = \sqrt{\frac{0.017098}{11}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7B0.017098%7D%7B11%7D%7D)
![s = \sqrt{0.00155436363}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B0.00155436363%7D)
-- approximated