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Serga [27]
3 years ago
7

Suppose unattached earlobes are a dominant trait. Phil and Maggie both have unattached earlobes but their daughter, Celia, does

not. If Phil and Maggie have a second child, what is the probability that he or she will have attached earlobes?
Mathematics
1 answer:
solong [7]3 years ago
4 0

Answer:

50/50

Step-by-step explanation: you can never tell really you and the husband or wife could both have blue eyes but your baby could have brown because your great great great grandma had them.

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ASAP - Need the answer and an explanation if possible!
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Answer:

It is 7,080

Step-by-step explanation:

Your trying to find the average speed therefore you divided 17,000 to 2.4 and after that as you said you round yo the nearst whole number and you get your answer.

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x > -2

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Lucy bought a 12 pound watermelon for $6.24.
Leokris [45]

Answer:

Part A: $0.52

Part B: $3.64

Step-by-step explanation:

Divide $6.24 by 12 pounds to determine the cost per pound

6.24/12=0.52

Multiply $0.52 by 7 pounds to find the cost for a 7-pound watermelon

0.52*7=3.64

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3 years ago
Can you please help me answer this picture​
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The following data summarizes results from 939 pedestrian deaths that were caused by accidents . If one of the pedestrian deaths
olganol [36]

Answer:

Step-by-step explanation:

Hello!

The contingency table is attached.

The total of pedestrians is 909, not 939, there are 30 accidents less in the given data.

I've calculated the asked probabilities using a total of 909.

1. The probability that the pedestrian was intoxicated or the driver was intoxicated.

Two events are mutually exclusive when the occurrence of one prevents the occurrence of the other in one single performance of the experiment. (i.e there is no intersection between the events P(A∩B)=0)

When two events are mutually exclusive, the probability of the union of both elements is equal to the sum of their individual probabilities P(A∪B)= P(A) + P(B).

When the events aren't mutually exclusive, the probability of their union is equal to the summary of their probabilities minus the probability of the intersection between these two events P(A∪B)= P(A) + P(B) - P(A∩B)

Where:

∪ is the union of both events, in colloquial language represents "or"

∩ is the union of both events, or intersection of the events, in colloquial language represents "and"

In this case P(P₁ ∪ D₁) = P(P₁) + P(D₁) - P(P₁ ∩ D₁) = \frac{306}{909} + \frac{132}{909} - \frac{81}{909} = 0.393

P₁ and D₁ are not mutually exclusive

2. The probability of the pedestrian was intoxicated or the driver was not intoxicated.

These two events aren't mutually exclusive.

P(P₁ ∪ D₂)= P(P₁) + P(D₂) - P(P₁ ∩ D₂)= \frac{306}{909} +\frac{777}{909} *-\frac{225}{909}= 0.944

I hope it helps!

3 0
3 years ago
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