2.32*103 = 2.39E3, or 2.39*10^3
This result is likely wrong because I first thought your 103 literally meant 103. Now it seems that you meant 10^3 (10 to the third power). So, ignore the above result and multiply again: 2.32*10^3. This is already in scientific notation.
4.2*10^4 is already in scientific notation.
4.2*10^(-2) is already in scientific notation.
5700=1.7*10^3 (answer in scientific notation)
6548000= 6.55*10^(-6) (in scientific notation)
Answer:
The correct answer is B. Contain generalizations and contractions.
Step-by-step explanation:
A generalization occurs when a specific idea is applied to a general category, in simple terms it occurs if a person states something always occur because this person saw it once. In this case, generalization occurs as this person affirms all stars are too far away and thus it will take a long time to get to those stars, however not all stars are as far away as this person believes and this does not mean it will take a long time to get there as new technology can be created, thus, this statement is not true and need revision. On the other hand, contractions refers to the process of shortening words by putting them together, this can be seen in "couldn't" ( could + not) and "they're" (they + are), contradictions are not correct as they should not be used in formal language and be avoided, thus, contradictions also need to be revised.
I would be 1034 because if you do 6x94 it’s 564 so if you do 11x94 you get the total witch is 1034
Answer:
a) P(B'|A) = 0.042
b) P(B|A') = 0.625
Step-by-step explanation:
Given that:
80% of the light aircraft that disappear while in flight in a certain country are subsequently discovered
Of the aircraft that are discovered, 63% have an emergency locator,
whereas 89% of the aircraft not discovered do not have such a locator.
From the given information; it is suitable we define the events in order to calculate the probabilities.
So, Let :
A = Locator
B = Discovered
A' = No Locator
B' = No Discovered
So; P(B) = 0.8
P(B') = 1 - P(B)
P(B') = 1- 0.8
P(B') = 0.2
P(A|B) = 0.63
P(A'|B) = 1 - P(A|B)
P(A'|B) = 1- 0.63
P(A'|B) = 0.37
P(A'|B') = 0.89
P(A|B') = 1 - P(A'|B')
P(A|B') = 1 - 0.89
P(A|B') = 0.11
Also;
P(B ∩ A) = P(A|B) P(B)
P(B ∩ A) = 0.63 × 0.8
P(B ∩ A) = 0.504
P(B ∩ A') = P(A'|B) P(B)
P(B ∩ A') = 0.37 × 0.8
P(B ∩ A') = 0.296
P(B' ∩ A) = P(A|B') P(B')
P(B' ∩ A) = 0.11 × 0.2
P(B' ∩ A) = 0.022
P(B' ∩ A') = P(A'|B') P(B')
P(B' ∩ A') = 0.89 × 0.2
P(B' ∩ A') = 0.178
Similarly:
P(A) = P(B ∩ A ) + P(B' ∩ A)
P(A) = 0.504 + 0.022
P(A) = 0.526
P(A') = 1 - P(A)
P(A') = 1 - 0.526
P(A') = 0.474
The probability that it will not be discovered given that it has an emergency locator is,
P(B'|A) = P(B' ∩ A)/P(A)
P(B'|A) = 0.022/0.526
P(B'|A) = 0.042
(b) If it does not have an emergency locator, what is the probability that it will be discovered?
The probability that it will be discovered given that it does not have an emergency locator is:
P(B|A') = P(B ∩ A')/P(A')
P(B|A') = 0.296/0.474
P(B|A') = 0.625
