Mathematics, 31.07.2020 02:01 dave92

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. A medical center observed that about 60% of its morning appointments were with elderly patients. The table shows the results of a simulation used to represent the scenario. The numbers 0 to 5 represent appointments with elderly patients, and the numbers 6 to 9 represent appointments with other patients.

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fra

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Answer from: ramzieboy13

1) 0.1504

2) 0.432

Step-by-step explanation:

1) The given information are;

The proportion of the morning appointments that are with elderly patients = 60%

The number of patients in the appointments = 10 patients

The proportion of the morning appointments that are with non-elderly patients = 100 - 60 = 40%

The binomial probability distribution is given as follows;

$P(X = r) = \dbinom{n}{r}p^{r}\left (1-p \right )^{n-r}$

$P(X = 0) = \dbinom{10}{0}0.6^{0}\left (1-0.6 \right )^{10}$ = 0.000105

$P(X = 1) = \dbinom{10}{1}0.6^{1}\left (1-0.6 \right )^{9}$ = 0.0016

$P(X = 2) = \dbinom{10}{2}0.6^{2}\left (1-0.6 \right )^{8}$ = 0.01062

$P(X = 3) = \dbinom{10}{3}0.6^{3}\left (1-0.6 \right )^{7}$ = 0.0425

The probability that the first four patients are elderly is 0.000105 + 0.0016 + 0.1062 + 0.0425 = 0.1504

2) The probability that exactly 2 out of 3 morning patients are elderly patient is given as follows

$P(X = 2) = \dbinom{3}{2}0.6^{2}\left (1-0.6 \right )^{1}$ = 0.432

Answer from: Quest

e) none of the above because the measure of angle cba is 101°

Find the measure of ∠cba. a) 25° b) 50° c) 75° d) 100° e) none of the above

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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