Hi there!
In order to answer your question, you'll need to use the cross product method (let's say that Newtons = N):
9N on Earth = 2N on Namar
400N on Earth = xN on Namar
(2 × 400) ÷ 9 = xN on Namar
800 ÷ 9 = xN on Namar
88.888.. = xN on Namar
Since your answer must be to the nearest Newton, I'm guessing that you need to round your answer to the nearest whole number. This means that since the number in the tenths column is more than 5 (could also be equal to 5), you need to round the number up.
88.888... rounded to the nearest whole number = 89
Your answer is: The girl weigh 89 newtons on Namar.
There you go! I really hope this helped, if there's anything just let me know! :)
Answer:
Step-by-step explanation:
hope you understand
Answer:
80
Step-by-step explanation:
78 is only 2 from 80 while it is 8 from 70
Answer:
1) is not possible
2) P(A∪B) = 0.7
3) 1- P(A∪B) =0.3
4) a) C=A∩B' and P(C)= 0.3
b) P(D)= 0.4
Step-by-step explanation:
1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4 . Thus the maximum possible value of P(A∩B) is 0.4
2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by
P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7
P(A∪B) = 0.7
3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3
4) the event C that the selected student has a visa card but not a MasterCard is given by C=A∩B' , where B' is the complement of B. Then
P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3
the probability for the event D=a student has exactly one of the cards is
P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4
Answer: a). 5
Explanation:
5 > 4
