Suppose the reaction temperature x (in °c) in a certain chemical process has a uniform distribution with a = −9 and b = 9. (a) compute p(x < 0). (no response) 0.5 (b) compute p(−4.5 < x < 4.5). (no response) 0.5 (c) compute p(−7 ≤ x ≤ 8). (round your answer to two decimal places.) (no response) 0.83 (d) for k satisfying −9 < k < k + 4 < 9, compute p(k < x < k + 4). (round your answer to two decimal places.) (no response) 0.22

A. P(x<0)=0,5

B. P(-4,5<x<4,5)=0,5

C. P(-7<x<8)=0,83

D. P(k<x<k+4)=0,22

Step-by-step explanation:

A. I hope that you need the reason why those answer are correct.

In a uniform distribution of probability, the probability of P(a<x<b) can be found as follows:

P(a<x<b)= (b-a)/B-A

where A,B are the limits of the function given A=-9, B=9. Because -9<x<9, then P(x<0)=P(-9>x>0) and now we use the formula:

P(-9<x<0)= 0-(-9)/( 9-(-9) )

P(-9<x<0)= 9/18 =0,5

B. We proceed using the formula as before:

P(-4,5<x<4,5)= (4,5-(-4,5)) / 18

P(-4,5<x<4,5)= 9/18 =0,5

C. And again, we use the formula:

P(-7<x<8)= (8-(-7)) / 18

P(-7<x<8)= 15/18 =0,8333... =0,83

D. This question could seem different because of the interval given there, let´s analyse it:

The inequality -9<k<k+4<9 it´s just a form to say that there is a probability for P(x<k) and P(x<k+4) because both k and k+4 belongs to the interval where the function is defined uniform. Now the question asks us, what is P(k<x<k+4), for all k that fills -9<k<k+4<9?

Because they fill -9<k<k+4<9, both of them belongs to [-9,9], then we can use the formula:

P(k<x<k+4)= k+4-(k) / 18

P(k<x<k+4)= 4/18 =0,2222... =0,22

21.4 i did it already for s1

step-by-step explanation:

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