After manipulating above equation we get, 2x^2 - 7x - 8= 0. Discriminant= b^2 - 4ac = (-7)^2 - 4(2)(-8) = 113>0. So there are 2 real roots :)
Answer:
<em>0</em> is the probability that a randomly selected student plays both a stringed and a brass instrument.
Step-by-step explanation:
Given that:
Number of students who play stringed instruments, N(A) = 15
Number of students who play brass instruments, N(B) = 20
Number of students who play neither, N(
)' = 5
<u>To find:</u>
The probability that a randomly selected students plays both = ?
<u>Solution:</u>
Total Number of students = N(A)+N(B)+N(
)' =15 + 20 + 5 = 40
(As there is no student common in both the instruments we can simply add the three values to find the total number of students)
As per the venn diagram, no student plays both the instruments i.e.
![N(A\cap B) =0](https://tex.z-dn.net/?f=N%28A%5Ccap%20B%29%20%3D0)
Formula for probability of an event E can be observed as:
![P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}](https://tex.z-dn.net/?f=P%28E%29%20%3D%20%5Cdfrac%7B%5Ctext%7BNumber%20of%20favorable%20cases%7D%7D%7B%5Ctext%20%7BTotal%20number%20of%20cases%7D%7D)
![P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%20%3D%20%5Cdfrac%7BN%28A%5Ccap%20B%29%7D%7BN%28U%29%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B0%7D%7B40%7D%5C%5C%5CRightarrow%20P%28A%5Ccap%20B%29%20%3D%200)
So, <em>0</em> is the probability that a randomly selected student plays both a stringed and a brass instrument.
850. If there is 17 red beads for every 150 beads Then you would divide 7500 by 150 and you get 50. Then multiply 50 with 17 and your answer is 850 red beads. 7500/150=50. 50 x 17= 850.
Answer:
yes
Step-by-step explanation: