Answer: 0.55567
Step-by-step explanation:
Given : A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.
i.e.
Let and are two randomly selected refrigerator's life whose sum will exceed third selected refrigerator .
So that,
Mean
Standard deviation =
Z-score :
Now , The probability that the total useful life of two(i..e n=2) randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator would be :-
Hence, the required probability is 0.55567.
Answer: <em>That is most likely an unsolvable question. </em>
Step-by-step explanation: PEMDAS
<u>3×4×100±67≤65</u> <em>Doesn’t work</em>.
<u>3x4= 12</u> ~ <u>12x100=1200</u> ~ <u>1200+67=1267</u> ~ 1267<65
556∪120D⊥p=13⇆87∞44f1······ <em>Also clearly isn’t a math question </em>
<em>I am not sure if this helps. But I really did look into it and found nothing. I am So sorry to disappoint</em>.
Answer:
r
Step-by-step explanation:
If signs are different you subtract. So 2.5- -1.25 equals 1.25 and keep the sign of the biggest number