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Butoxors [25]
3 years ago
7

Consider the following experiment: triplets are born and the order of birth of boys and girls is recorded.

Mathematics
2 answers:
Scrat [10]3 years ago
8 0
No, it is not the compliment.

If you have two complimentary events, they account for all possible outcomes of a situation. With these two events, we don't account for having exactly 2 children that are boys.
SOVA2 [1]3 years ago
4 0

Answer:

No, event B is not complement of event A.

Step-by-step explanation:

Event B is not a compliment of Event A because in complement we consider all those outcomes of the event which we don't want.If those outcomes match with the outcomes of the other event then we consider it as a complement of other event.Here in this case the outcomes which we don't want are not the complement of the outcomes of the other event.


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She receives $1,102.50 for commission. All you need to do to solve this problem is multiply the amount given by the percentage as a decimal.
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2 years ago
2(3x + 5) - 16 + 2<br> due soon
Zanzabum
PEMDAS

multiply 2 in the parentheses:

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2 years ago
A 3-column table with 4 rows. Column 1 is labeled Inequality with entries 8 greater-than 2, negative 10 less-than negative 2, 15
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4 0
3 years ago
Find the sum of the first 47 terms of the following series, to the nearest integer.
Leokris [45]

Answer:

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ is 4,888

Step-by-step explanation:

The given series is;

12, 16, 20, ...,

Therefore, the first term of the series is, a = 12

The common difference of series is found as follows;

The difference between subsequent terms, 12 and 16 is  16 - 12 = 4

The difference between subsequent terms, 16, and 20 is  20 - 16 = 4

Therefore, the common difference, d = 4

The series is therefore an arithmetic projection, AP

The sum of the first 'n' terms of an AP, Sₙ, is given as follows;

S_n = \dfrac{n}{2} \cdot \left [2 \cdot a + (n - 1)\cdot d \right ]

(47/2)*(2*12+(47-1)*4)

The sum of the first 47 terms is therefore given as follows;

S_n = \dfrac{47}{2} \cdot \left [2 \times 12 + (47 - 1)\times 4 \right ] = 4,888

The sum of the first 47 terms of the series, 12, 16, 20, ... S₄₈ = 4,888

5 0
2 years ago
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