The sum of the two prime numbers is 85. This is an odd number. According to the Rule, the sum of the even prime number and the odd prime number is called an odd number.
The thing is that every 4 years, the tree grows 6 inches, so that could be expressed as a ratio, 4/6. That also means that every two years, the tree grows 3 inches since it could be simplified to 2/3.
<h3>
Answer: 80 meters</h3>
This is an isosceles triangle. The dashed line is the height which is perpendicular to the base 120. The height is always perpendicular to the base. The dashed line cuts the base into two equal pieces (this only works for isosceles triangles when you cut at the vertex like this).
So we have two smaller triangles each with a base of 60 and a height of x. Focus on one of the right triangles and use the pythagorean theorem to solve for x.
a^2 + b^2 = c^2
x^2 + (60)^2 = (100)^2
x^2 + 3600 = 10000
x^2 = 10000 - 3600
x^2 = 6400
x = sqrt(6400)
x = 80
Each smaller right triangle has side lengths of 60, 80, 100
Note the ratio 60:80:100 reduces to 3:4:5. A 3-4-5 right triangle is a very common pythagorean primitive.
Answer:
x = 3.51
Step-by-step explanation:
Since, formula to determine the area f a triangle is,
Area = 
8 = 
16 = x(3x - 7) + 1(3x - 7)
16 = 3x² - 7x + 3x - 7
16 = 3x² - 4x - 7
0 = 3x² - 4x - 23
3x²- 4x - 23 = 0
By quadratic formula,
x = 
x = 
x = 
x = 
x = 3.51, -2.18
But the length of sides can't be negative.
Therefore, x = 3.51 will be the answer.
Answer:
Probability of stopping the machine when 
is 0.0002
Probability of stopping the machine when 
is 0.0013
Probability of stopping the machine when 
is 0.0082
Probability of stopping the machine when 
is 0.0399
Step-by-step explanation:
There is a random binomial variable 
that represents the number of units come off the line within product specifications in a review of 
Bernoulli-type trials with probability of success 
. Therefore, the model is 
. So:
![P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%209%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%209%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%209%7D%20%280.91%29%5E%7B9%7D%280.09%29%5E%7B6%7D%2B...%2B%7B%2015%20%5Cchoose%2015%7D%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0002%20)
![P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2010%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2010%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2010%7D%280.91%29%5E%7B10%7D%280.09%29%5E%7B5%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0013%20)
![P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2011%29%20%3D%201%20-%20P%20%28X%20%5Cgeq%2011%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2011%7D%280.91%29%5E%7B11%7D%280.09%29%5E%7B4%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0082)
![P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399](https://tex.z-dn.net/?f=%20P%20%28X%20%3C%2012%29%20%3D%201-%20P%20%28X%20%5Cgeq%2012%29%20%3D%201%20-%20%5B%7B15%20%5Cchoose%2012%7D%280.91%29%5E%7B12%7D%280.09%29%5E%7B3%7D%2B...%2B%7B15%20%5Cchoose%2015%7D%20%280.91%29%5E%7B15%7D%280.09%29%5E%7B0%7D%5D%20%3D%200.0399%20)
Probability of stopping the machine when is 0.0002
Probability of stopping the machine when is 0.0013
Probability of stopping the machine when is 0.0082
Probability of stopping the machine when is 0.0399
