The correct answer to this question is 0.34 or 34 percent. To solve this, use the formula P(nuts/chocolate) = P(chocolate and nuts)/P(chocolate) = 0.25/0.73 = 0.342. Round to the nearest tenth of a percent, the answer is 34 percent. This is <span>the probability that a dessert chosen at random contains nuts given that it contains chocolate. </span>Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Answer:
try 65 degrees
Step-by-step explanation:
Answer:
86
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that total students =23.
15 have taken English, 13 Math and 8 both
number of students taking math only = total math - students who have taken both
= 13-8 = 5
Number of students taking English only = total English - students who have taken both
= 15-8 =7
Now we divided into 3 disjoint partitions as only Math, only English and both.
Thus no of students who have taken either math or english = 7+5+3 = 15
No of students who have taken neither = 24-15 = 9
Answer:
ITS C
Step-by-step explanation:
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