Answer:
Step-by-step explanation:
a. when t=0,h=3ft
![h(t)=-16 (t^2-\frac{47}{16}x+(\frac{47}{32} )^2)+\frac{47^2}{32^2} *16+3\\=-16(t-\frac{47}{32})^2+\frac{2209}{64}+3\\=-16(t-\frac{47}{32})^2+\frac{2209+192}{64}\\when~ t=\frac{47}{32}\\h=\frac{2301}{64}\\or~v^2-u^2=2gh\\\\0^2-47^2=2*(-32)h\\or h=\frac{47^2}{64}=\frac{2209}{64} \\height from the ground=\frac{2209}{64} +3=\frac{2301}{64}](https://tex.z-dn.net/?f=h%28t%29%3D-16%20%28t%5E2-%5Cfrac%7B47%7D%7B16%7Dx%2B%28%5Cfrac%7B47%7D%7B32%7D%20%29%5E2%29%2B%5Cfrac%7B47%5E2%7D%7B32%5E2%7D%20%2A16%2B3%5C%5C%3D-16%28t-%5Cfrac%7B47%7D%7B32%7D%29%5E2%2B%5Cfrac%7B2209%7D%7B64%7D%2B3%5C%5C%3D-16%28t-%5Cfrac%7B47%7D%7B32%7D%29%5E2%2B%5Cfrac%7B2209%2B192%7D%7B64%7D%5C%5Cwhen~%20t%3D%5Cfrac%7B47%7D%7B32%7D%5C%5Ch%3D%5Cfrac%7B2301%7D%7B64%7D%5C%5Cor~v%5E2-u%5E2%3D2gh%5C%5C%5C%5C0%5E2-47%5E2%3D2%2A%28-32%29h%5C%5Cor%20h%3D%5Cfrac%7B47%5E2%7D%7B64%7D%3D%5Cfrac%7B2209%7D%7B64%7D%20%5C%5Cheight%20from%20the%20ground%3D%5Cfrac%7B2209%7D%7B64%7D%20%2B3%3D%5Cfrac%7B2301%7D%7B64%7D)
c.t=47/32 sec
d.
time taken to fall 2301/64 ft=0*t+\frac{1}{2}*32*t^2
t^2=\frac{2301}{64*16}
t=\frac{\sqrt{2301}}{8*4}=\frac {47.97}{32}
total ~time=\frac{47+47.97}{32}=\frac{94.97}{32} sec
![\mathbb P(Z](https://tex.z-dn.net/?f=%5Cmathbb%20P%28Z%3Cz%29%3DF_Z%28z%29%3D0.628%5Cimplies%20z%3D%7BF_Z%7D%5E%7B-1%7D%280.628%29%5Capprox0.3266)
where
![F_Z(z)](https://tex.z-dn.net/?f=F_Z%28z%29)
is the cumulative distribution function for the random variable
![Z](https://tex.z-dn.net/?f=Z)
following the standard normal distribution.
It is 2.4 cups of 10% fruit juice and berry punch and 9.6 cups of 20%fruit juice:
let x to be 10% fruit juice and berry punch and y= 20% fruit juice
.x+.2y=2.16
x+y=2.4
use substitution property or elimination property to solve the problem:
x=2.4 and y=9.6
suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = ![\frac{x-mean}{standard deviation}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx-mean%7D%7Bstandard%20deviation%7D%20%20)
= ![\frac{174-177}{26}](https://tex.z-dn.net/?f=%20%5Cfrac%7B174-177%7D%7B26%7D%20%20)
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438