We have equation:
p = number of phones left to repair
108 = initial amount at the start of week
23 = number of phones repaired each day
d = number of days she worked that week
1.) 17.1g > 1.71mg
2.) 6.3cm< 63m
3.)1250ml>12.5
4.) 7/12 < 2/3
5.) 7/10 < 11/15
Answer:
0.1319 or 13.2%
Step-by-step explanation:
You can solve this using the binomial probability formula.
The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.
Then, we can set the equation as follows:
P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k)
n=4, x=2, k=2
when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157
when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154
when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008
Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)
All you need is use sin cos tan for doing this
sin (angle)= O/H
cos (angle)= A/H
tan (angle)= O/A
