Answer:
Tha answer of this question is an angle
Answer:
40
Step-by-step explanation:
D=2r so 2*20=40
Answer:
One solution
Step-by-step explanation:
As a generalization: an equation of degree n has n solutions. A quadratic equation has the second degree and thus two a solutions; a third degree equation has three, and so on. Thus, a first degree (linear) equation has one solution.
Let as consider the complete ques is : Ray QS bisects ∠PQR. Solve for x and find m∠PQR. m∠PQS = 3x ; m∠SQR = 5x-20.
Given:
Ray QS bisects ∠PQR.
m∠PQS = 3x
m∠SQR = 5x-20
To find:
The value of x and m∠PQR
Solution:
Ray QS bisects ∠PQR. So,





The value of x is 10.
Now,



Put x=10,



Therefore, the m∠PQR is 60 degrees.
The option fourth "There is a strong negative relationship between x and y" is correct.
<h3>What is correlation?</h3>
It is defined as the relation between two variables which is a quantitative type and gives an idea about the direction of these two variables.
![\rm r = \dfrac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{{[n\sum x^2- (\sum x)^2]}}\sqrt{[n\sum y^2- (\sum y)^2]}}](https://tex.z-dn.net/?f=%5Crm%20r%20%3D%20%5Cdfrac%7Bn%28%5Csum%20xy%29-%28%5Csum%20x%29%28%5Csum%20y%29%7D%7B%5Csqrt%7B%7B%5Bn%5Csum%20x%5E2-%20%28%5Csum%20x%29%5E2%5D%7D%7D%5Csqrt%7B%5Bn%5Csum%20y%5E2-%20%28%5Csum%20y%29%5E2%5D%7D%7D)
The options are:
- There is no relationship between x and y.
- There is a weak negative relationship between x and y.
- There is a strong positive relationship between x and y.
- There is a strong negative relationship between x and y.
We have data:
x 7 8 9 10 11 12 13
y 13 10 7 6 5 3 0
The value of correlation r = -0.9847
As we know the value of r between -1 to +1
If r = 1 (perfect and positive correlation)
if r = -1 (perfect and negative correlation)
We have r = -0.9847
The value is near to -1 which means there is a strong negative relationship between x and y.
Thus, the option fourth "There is a strong negative relationship between x and y" is correct.
Learn more about the correlation here:
brainly.com/question/11705632
#SPJ1