I believe The answer is B
To find the volume of this one we need to break it down
now i see half of a cylinder and rectangle:)
but first lets find the volume of the rectangle...
In order to find the Volume of a rectangle we need to use this formula...
Length x width x height
in this case...
length = 10in
width = 6 in
height = 8in
lets solve:)
10 x 6 x 8 = 480
or we write it like this
480in³
now time to find the volume of the half cylinder:)
But first lets remember the volume for a cylinder
Volume =

So lets find our measurements

= 3.14
r² = 5² or 25
h = 6
so lets plug in our values just like our formula said:)
3.14 x 25 x 6
now lets easily solve
<span>3.14 x 25 x 6 = 471
</span>now since we found an entire cylinder and we only want half of a cylinder lets divide our answer in half
471 ÷ 2 = 235.5
so we write it like this 235.5units³
But we have to add both of our multiples together so lets do that
Volume of rectangle = <span>480in³
</span>volume of half sphere = 235.5units³
480 + 235.5 = 715.5
answer = 715.5units³
I hope this helped and everyone learned something new
anyways don't forget to
MARK ME BRAINLIEST! :D
Answer: If you see this line solid its > or equal too this > with line underneath
If you see line broken it's only >
Y>_ - 1/4x+2
Step-by-step explanation:
Starts at 2 on y then next pt go down 1 and 4 to the right
Answer:
60 divided 25
Step-by-step explanation:
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)