since we have the area of the front side, to get its volume we can simple get the product of the area and the length, let's firstly change the mixed fractions to improper fractions.
![\stackrel{mixed}{23\frac{2}{3}}\implies \cfrac{23\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{71}{3}} ~\hfill \stackrel{mixed}{4\frac{7}{8}}\implies \cfrac{4\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{39}{8}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{71}{3}\cdot \cfrac{39}{8}\implies \cfrac{71}{8}\cdot \cfrac{39}{3}\implies \cfrac{71}{8}\cdot 13\implies \cfrac{923}{8}\implies 115\frac{3}{8}~in^3](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B23%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B23%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B71%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B7%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%208%2B7%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B39%7D%7B8%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B71%7D%7B3%7D%5Ccdot%20%5Ccfrac%7B39%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B71%7D%7B8%7D%5Ccdot%20%5Ccfrac%7B39%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B71%7D%7B8%7D%5Ccdot%2013%5Cimplies%20%5Ccfrac%7B923%7D%7B8%7D%5Cimplies%20115%5Cfrac%7B3%7D%7B8%7D~in%5E3)
If it is geometric then
if the terms are a,b,c,d then
b/a=c/b=d/c
so
-54/-81=-36/-54=-24/-36?
2/3=2/3=2/3
and that is the common ratio
it is geomertic and common ratio is 2/3
He drank 420 cups of water in one month
Surface area of box=1200 cm²
<span>Volume of box=s²h </span>
<span>s = side of square base </span>
<span>h = height of box </span>
<span>S.A. = s² + 4sh </span>
<span>S.A. = surface area or 1200 cm², s²
= the square base, and 4sh = the four 'walls' of the box. </span>
<span>1200 = s² + 4sh </span>
<span>1200 - s² = 4sh </span>
<span>(1200 - s²)/(4s) = h </span>
<span>v(s) = s²((1200 - s²)/(4s)) </span>
<span>v(s) = s(1200 - s²)/4 . </span>
<span>v(s) = 300s - (1/4)s^3</span>
by derivating
<span>v'(s) = 300 - (3/4)s² </span>
<span>0 = 300 - (3/4)s² </span>
<span>-300 = (-3/4)s² </span>
<span>400 = s² </span>
<span>s = -20 and 20. </span>
again derivating
<span>v"(s) = -(3/2)s </span>
<span>v"(-20) = -(3/2)(-20) </span>
<span>v"(-20) = 30 </span>
<span>v"(20) = -(3/2)(20) </span>
<span>v"(20) = -30 </span>
<span>v(s) = 300s - (1/4)s^3 </span>
<span>v(s) = 300(20) - (1/4)(20)^3 </span>
<span>v(s) = 6000 - (1/4)(8000) </span>
<span>v = 6000 - 2000
v=4000</span>
Answer:
Option D) $275
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $235
Standard Deviation, σ = $20
We are given that the distribution of amount of money spent by students is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.975
Calculation the value from standard normal z table, we have,
Approximately 97.5% of the students spent below $275 on textbook.
