Answer:
109
Step-by-step explanation:
You can brute-force this:
Candidate numbers that are divisable by 13 if you take away 5: 18
, 31
, 44
, 57
, 70
, 83
, 96
, <u>109
</u>, 122
, 135
, 148
Candidate numbers that are divisable by 14 if you take away 11: 25
, 39
, 53
, 67
, 81
, 95
, <u>109
</u>, 123
, 137
Only 109 is in both lists.
Total possible outcomes: (2)³=8 (heads or tails possibility on each toss)
Draw out a Tree diagram to illustrate the given example, labelling the probability on each branch.
Adding all the possible situations with P(t)≥2, we get 0.352 as the probability.
Hope I helped :)
1700 because I multiply 30 x 20 then 20 x 5 then 30 x 5 then add them all up and multiply your answer by 2
Hi! Let me help you!
- Remember, factoring looks like so:
- ab+ac=a(b+c)
- So, we need to find the greatest common factor (G.C.F.) of the polynomial.
- In this case, 6 is the GCF, so we factor it out:
- 6x²+66x+60
- Divide 6, 66 and 60 by 6:
- x²+11x+10
<u>Answer:</u>
x²+11x+10
Hope you find it helpful.
