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Leni [432]
3 years ago
15

question 1 a physician’s “patient panel” is the list of all patients under their care. in the us, the size of patient panels amo

ng primary care doctors is well modeled by a n (1500, 200) distribution. (a) draw a normal curve for this model, labeling the empirical rule (68-95-99.7 rule). (b) if a “medium-sized” panel has anywhere from 1100 to 1700 patients, what proportion of us primary care doctors have a patient panel that is not “medium-sized”? (c) find q1,q3 and the iqr for this distribution.
Mathematics
1 answer:
Molodets [167]3 years ago
7 0

Answer:

sorry but i really do’t know

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I'll look into that answer

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Find the perimeter.
Scorpion4ik [409]

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7 1/3 inches

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2 (1 3/4 + 1 11/12)

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4 0
3 years ago
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babunello [35]

Answer: -40 or negative forty

Step-by-step explanation:

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In a bag of m&m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\
=1140

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=
\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
4 0
3 years ago
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