Jodi poured herself a cold soda that had an initial temperature 36 degrees f and immediately went outside to sunbathe where the temperature was a steady 99 degree f. after 5minutes the temperature of the soda was 46 degree f .jodi had to run back into the house to answer the phone . what is the expected temperature of the soda after an additional 13 minutes ?

this can be solve using newtons heating of cooling

(Ts – T) =(Ts – To)*e^(-kt)

Where Ts is the ambient temperature

To is initial temperature

T is the temperature at time t

t is the time

k is constant

fisrt solve the constant k for the given first scenario

(99 – 36) = (99 – 46)*e(-5k)

K = -0.0346

Using k, solve T at t = 13 min

(99 – 46) = (99 – T)*e(-13*(-0.0346)

T = 58.82 degree F

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

walter needs to make at least 5 more cookies but no more than 40. 5 ≥ x ≥ 40;

step-by-step explanation:

walter has to make a minimum of 20 cookies. he already has 15 cookies made, so he must make 5 more to meet the minimum.

the maximum amount of cookies he can make is 55. he already has 15, so he has to make no more than 40 more cookies to meet the maximum.