The pair of numbers is (3,3) while the maximum product is 9
Step-by-step explanation:
The pairs of numbers whose sum is 6 starting from zero is ;
0,6
1,5
2,4
3,3
Kindly note 2,4 is same as 4,2 , so there is no need for repetition
So the maximum product is 3 * 3 = 9 and the pair is 3,3
Answer: ![\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}](https://tex.z-dn.net/?f=%5Cbold%7B5%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Csqrt%7B11%7D%7D%7B6%7D%5Cqquad%206%29%5C%20%5Ctan%20%5Ctheta%20%3D%5Cdfrac%7B8%7D%7B17%7D%5Cqquad%207%29%5C%20%5Ccos%20%5Ctheta%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%5Cqquad%208%29%5C%20%5Ccos%20%5Ctheta%20%3D%20%5Cdfrac%7B%5Csqrt%7B10%7D%7D%7B10%7D%7D)
<u>Step-by-Step Explanation:</u>
Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>
![5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}](https://tex.z-dn.net/?f=5%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B3%5Csqrt%7B11%7D%7D%7B18%7D%5Cquad%20%5Crightarrow%20%5Clarge%5Cboxed%7B%5Cdfrac%7B%5Csqrt%7B11%7D%7D%7B6%7D%7D)
Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18
![6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}](https://tex.z-dn.net/?f=6%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B8%7D%7B17%7D%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B8%7D%7B17%7D%7D)
Note: 8² + 15² = hypotenuse² → hypotenuse = 17
![7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}](https://tex.z-dn.net/?f=7%29%5C%20%5Ctan%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20opposite%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%3D%5Cdfrac%7B20%7D%7B15%7D%5Cquad%20%5Crightarrow%20%5Clarge%5Cboxed%7B%5Cdfrac%7B4%7D%7B3%7D%7D)
Note: hypotenuse not needed for tan
![8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}](https://tex.z-dn.net/?f=8%29%5C%20%5Ccos%20%5Ctheta%3D%5Cdfrac%7B%5Ctext%7Bside%20adjacent%20to%7D%5C%20%5Ctheta%7D%7B%5Ctext%7Bhypotenuse%20of%20triangle%7D%7D%3D%5Cdfrac%7B2%7D%7B2%5Csqrt%7B10%7D%7D%5Cquad%20%3D%5Clarge%5Cboxed%7B%5Cdfrac%7B%5Csqrt%7B10%7D%7D%7B10%7D%7D)
Note: 2² + 6² = hypotenuse² → hypotenuse = 2√10
Answer:
0.00665
0.006951
Step-by-step explanation:
If the first child is born with the affliction, there's a 0.05 probability that the second child will be born with the affliction. Which means there's a 0.95 probability that they won't.
The probability that the first child is born with the affliction but not the second is therefore:
P = 0.007 × 0.95
P = 0.00665
There's a 0.007 probability that the first child will be born with the affliction, which means there's a 0.993 probability that they won't. If the first child isn't born with the affliction, then the probability the second child will have it is 0.007.
So the probability that the first child is not born with the affliction and the second child is would be:
P = 0.993 × 0.007
P = 0.006951
Since the exponent is on the outside of the paranthesis, you'd have to multiply it by everything on the inside: a^m x b^m
1 = 27
2 = 27
3 = 27
4 = 27
5 = 126