Answer:
B . Yes
Step-by-step explanation:
Recall: a tangent of a circle is perpendicular to the radius of a circle, forming a right angle at the point of tangency.
Therefore, if segment ST is tangent to circle P, it means that m<T = 90°, and ∆PST is a right triangle.
To know if ∆PST is a right triangle, the side lengths should satisfy the Pythagorean triple rule, which is:
c² = a² + b², where,
c = longest side (hypotenuse) = 37
a = 12
b = 35
Plug in the value
37² = 12² + 35²
1,369 = 1,369 (true)
Therefore we can conclude that ∆PST is a right triangle, this implies that m<T = 90°.
Thus, segment ST is a tangent to circle P.
Answer:
z = 2.1784 > 1.96,
Reject the null hypothesis
Step-by-step explanation:
For the males:
n1 = 162, x1 = 63
P1 = x 1/ n1 = 0.3889
For the Females:
n2 = 333,
x2 = 97
P 2 = x2/n2
= 0.2913
P 1= P2 Null hypothesis
P 1 is not equal to P 2 alternative hypothesis
Pooled proportion:
P= (x1 + x2) /( n1+ n2)
= (63 + 97) / (162 + 333)= 0.3232
Test statistics :
Z= (p1 - p2) /√p(1-p)× (1/n1 + 1/n2)
0.3889- 0.2913 / √0.3232 × 0.6768 × (1/162 +1/333)
=2.1784
c) Critical value :
Two tailed critical value, z critical = Norm.S .INV (0.05/2) = 1.960
Reject H o if z < -1.96 or if z > 1.96
d) Decision:
z = 2.1784 > 1.96,
Reject the null hypothesis
Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
