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Xelga [282]
3 years ago
10

Find the probability of at least 2 girls in births. Assume that male and female births are equally likely and that the births ar

e independent events. Round to three decimal places.
Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0
We assume that male and female births are equally likely, it means that the probability of birth of male= probability of birth of female = 100%/2=50% or 0.5We have 2 independent events. So what will be the variants:
Male and FemaleMale  and Male Female and Male Female and Female 
All four variants are equally likely.Probability of each one is 1/4 = 0.25.So, Result "Female and Female"
 are probability = 0.250 --- if round to three decimal places.
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Let the random variable X denote the number of network blackouts in a day. The
faltersainse [42]
E(X) = 0(0.7) + 1(0.2) + 2(0.1) = 0.2 + 0.2 = 0.4
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3 years ago
What percent of 11 is 44%
Inga [223]

Answer:

4

Step-by-step explanation:

cause math

3 0
3 years ago
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Aleksandr [31]

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2599.50

Step-by-step explanation:

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3 years ago
Find the measure of ZDCF. C F mDF = 49 G E MEG = 127 Find the measure of angle DCF​
AleksandrR [38]

Answer:

Step-by-step explanation:

Comment

If two secants intersect outside a circle (as these two do) then the angle at which they meet is 1/2 the difference between the intersected arcs. Put much simpler <DCF = 1/2 (arc EG - arc DF)

Givens

Arc EG = 127

Arc DF = 49

Solution

<DCF = 1/2(127 - 49)

<DCF = 1/2(78)

<DCF = 39

8 0
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sergey [27]

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Reduce & Cancel: 165°

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