Answer:
First score is 90
Step-by-step explanation:
Let A represent the first score
Let B represent the second score
Let C represent the third score
A+B+C=217 equation 1
If the first score A is 30 points more than the second,B
A=B-30 equation 2
Lastly,sum of A&B is 16 more than 2C
A+B=2C+16 equation 3
From equation 1
A+B=217-C equation 4
substitute for A+B in equation 3
217-C=2C+16
217-16=2C+C
201=3C
C=67
substituting C in equation 4
A+B=217-67
A+B=150
if the first score more than the first,then the first 90 while second 60 since A+B=150
Answer:
a) 
b) 
And replacing we got:
![P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%205%29%20%3D%201-%5B0.01%2B0.02%2B0.03%2B0.12%2B0.11%5D%3D1-0.29%3D0.71)
Step-by-step explanation:
For this case we can solve this problem creating the following table
Number of particles Frequency Rel. Frequency
0 1 1/100 =0.01
1 2 2/100 =0.02
2 3 3/100=0.03
3 12 12/100=0.12
4 11 11/100=0.11
5 15 15/100=0.15
6 18 18/100=0.18
7 10 10/100=0.1
8 12 12/100=0.12
9 4 4/100=0.04
10 5 5/100=0.05
11 3 3/100=0.03
12 1 1/100=0.01
13 2 2/100=0.02
14 1 1/100=0.01
Total 100 1
We assume on this case the the relative frequency represent the probability.
Let X the number of contaminating particles on a silicon wafer
What proportion of the sampled wafers had at least one particle?
For this case we can use the complement rule like this:

At least five particles?
Again for this case we can use the complement rule and we got:

And replacing we got:
![P(X \geq 5) = 1-[0.01+0.02+0.03+0.12+0.11]=1-0.29=0.71](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%205%29%20%3D%201-%5B0.01%2B0.02%2B0.03%2B0.12%2B0.11%5D%3D1-0.29%3D0.71)
Answer:
43,046,721
Step-by-step explanation:
Solve in the parentheses then go to the brackets.
3 to the 2nd power is 9
9 to the second power is 81
81 to the 4th power is 43,046,721.