The domain would be (4, infinity) so most likely your answer would be D.
Answer:
x=mc2 + 86
Step-by-step explanation:
Answer:
The probability that 13 of them were very confident their major would lead to a good job is 1.08%.
Step-by-step explanation:
Given : A 2017 poll found that 56% of college students were very confident that their major will lead to a good job. If 15 college students are chosen at random.
To find : What's the probability that 13 of them were very confident their major would lead to a good job?
Solution :
Applying Binomial distribution,
![P(x)=^nC_x p^x q^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3D%5EnC_x%20p%5Ex%20q%5E%7Bn-x%7D)
Here, p is the success p=56%=0.56
q is the failure ![q= 1-p=1-0.56=0.44](https://tex.z-dn.net/?f=q%3D%201-p%3D1-0.56%3D0.44)
n is the number of selection n=15
The probability that 13 of them were very confident their major would lead to a good job i.e. x=13
Substitute the values,
![P(13)=^{15}C_{13} (0.56)^{13} (0.44)^{15-13}](https://tex.z-dn.net/?f=P%2813%29%3D%5E%7B15%7DC_%7B13%7D%20%280.56%29%5E%7B13%7D%20%280.44%29%5E%7B15-13%7D)
![P(13)=\frac{15!}{13!2!}\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D%5Cfrac%7B15%21%7D%7B13%212%21%7D%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=\frac{15\times 14}{2\times 1}\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D%5Cfrac%7B15%5Ctimes%2014%7D%7B2%5Ctimes%201%7D%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=105\times (0.56)^{13}\times (0.44)^{2}](https://tex.z-dn.net/?f=P%2813%29%3D105%5Ctimes%20%280.56%29%5E%7B13%7D%5Ctimes%20%280.44%29%5E%7B2%7D)
![P(13)=0.0108](https://tex.z-dn.net/?f=P%2813%29%3D0.0108)
The probability that 13 of them were very confident their major would lead to a good job is 1.08%.
X=1 Y=4 because 32-20 = 12, so 12/3 = 4.
5x4=20 20+(-15) = 5. 5/5=1.