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nirvana33 [79]
3 years ago
5

Of the 49 plays attributed to a​ playwright, 18 are​ comedies, 19 are​ tragedies, and 12 are histories. If one play is selected

at​ random, find the odds against selecting a histories or a comedy
Mathematics
1 answer:
saul85 [17]3 years ago
6 0

Answer:

34:49

Step-by-step explanation:

total = 49 plays

comedy + tragedies is 20 + 14 = 3

So the odds of picking a tragedy or comedy is 34:49

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ss7ja [257]
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3 0
2 years ago
For i≥1 , let Xi∼G1/2 be distributed Geometrically with parameter 1/2 . Define Yn=1n−−√∑i=1n(Xi−2) Approximate P(−1≤Yn≤2) with l
murzikaleks [220]

Answer:

The answer is "0.68".

Step-by-step explanation:

Given value:

X_i \sim \frac{G_1}{2}

E(X_i)=2 \\

Var (X_i)= \frac{1- \frac{1}{2}}{(\frac{1}{2})^2}\\

             = \frac{ \frac{2-1}{2}}{\frac{1}{4}}\\\\= \frac{ \frac{1}{2}}{\frac{1}{4}}\\\\= \frac{1}{2} \times \frac{4}{1}\\\\= \frac{4}{2}\\\\=2

Now we calculate the \bar X \sim N(2, \sqrt{\frac{2}{n}})\\

\to \frac{\bar X - 2}{\sqrt{\frac{2}{n}}}  \sim  N(0, 1)\\

\to \sum^n_{i=1}  \frac{X_i - 2}{n}  \times\sqrt{\frac{n}{2}}}  \sim  N(0, 1)\\\\\to  \sum^n_{i=1}  \frac{X_i - 2}{\sqrt{2n}}  \sim  N(0, 1)\\

\to Z_n = \frac{1}{\sqrt{n}} \sum^n_{i=1} (X_i -2) \sim N(0, 2)\\

\to P(-1 \leq X_n \leq 2)  = P(Z_n \leq Z) -P(Z_n \leq -1) \\\\

                               = 0.92 -0.24\\\\= 0.68

6 0
3 years ago
Q: how do I find the average of a line plot?<br><br><br><br> A: look at the picture
ivolga24 [154]
You just answerd it yourself i think.......
6 0
3 years ago
Suppose the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average
arsen [322]

Answer:

0.5372

Step-by-step explanation:

Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.

Let X be the no of births in the hospital per hour

X is Poisson

with mean = 1.8

the probability of observing at least two births in a given hour at the hospital

= P(X\geq 2)\\= 1-F(1)\\= 1-0.4628\\= 0.5372

the probability of observing at least two births in a given hour at the hospital = 0.5372

4 0
3 years ago
I need help on 6-98 can you please help me
OLga [1]
Molly got 20 because she did not follow the order of operations, so this is incorrect. Nancy got the correct answer because she multiplied 4 * 3, and then added 2, giving her 14. Molly added 3+2, and multiplied that by 4. Nancy is correct with 14.


4 0
3 years ago
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