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alexgriva [62]
2 years ago
13

If A and B are independent events with P(A) = 0.5 and P(B) = 0.5, then P(A ∩ B) a. is 1.00. b. is 0.5. c. is 0.00. d. None of th

ese alternatives are correct.
Mathematics
1 answer:
erastova [34]2 years ago
3 0

Answer:

The answer is (d) "None of the these alternatives are correct"

Step-by-step explanation:

If two events A and B are independent, the probability of the intersection P(A\cap B) is defined as:

P(A\cap B)=P(A)\cdot P(B)

Therefore, in the exercise:

P(A\cap B)=P(A)\cdot P(B)=0.5\cdot 0.5\\P(A\cap B)=0.25

which give (d) "None of the these alternatives are correct"

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<h3>How many chocolate ice creams are sold?​</h3>

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1 year ago
What is the average cost of 7 articles if 3 of them cost 30k each and the rest cost 12.5k each ​
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{\bold{\red{\huge{\mathbb{QUESTION}}}}}

what is the average cost of 7 articles if 3 of them cost 30k each and the rest cost 12.5k each

\bold{ \red{\star{\blue{GIVEN }}}}

TOTAL NUMBER OF ARTICLE= 7

ARTICLES FOR 30K = 3

ARTICLES FOR 12.5K = 7-3= 4

\bold{\blue{\star{\red{TO \:  \: FIND}}}}

Average cost of 7 articles

\bold{  \green{ \star{ \orange{FORMULA \:  USED}}}}

\red{AVERAGE  =  \frac{TOTAL \:  COST}{ NUMBER  \: OF\: ARTICLES }}

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Average \:  cost  \: of  \: 7  \: articles -  >  \\  \frac{(3 \times 30k) + (12 .5k \times 4)}{7}  \\  \frac{90k + 50k}{7}  \\  \frac{140k}{7}  \\  \blue {Average \:  cost  \: of  \: 7  \: articles -  >} \\ 20k

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