Give an example of (y,m’) a measurable space
F : X⇒Y
M = { E ⊆ X : F(E) ∈ Y }
M is not a sigma_algebra on x
Answers
Step-by-step explanation:
- The F1 = {∅,Ω,{a}} is not an algebra or σ-algebra, but F2 = {∅,Ω,{a},{b, c, d}} is a σ-algebra. Example 2 Let Ω be an infinite set. Then F5, the collection of all subsets of Ω that are either finite sets or have complements that are finite sets, is an algebra but not a σ-algebra.
Answer:
Given : A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Need to Find : The volume of wood and graphite
______________________________________
★ Cᴏɴᴄᴇᴘᴛ :
According to the question, here we are provided with pencil and graphite which has a diameter of 7mm and 1mm. Now, as we know that graphite in always placed inside the pencil which helps us to write something. Now, the length of the pencil is 14 cm. First we need to get the volume of pencil using the formula for getting the volume of cylinder as a pencil is cylindrical in shape. The. using the same formula we can get the volume of wood. After which subtracting the volume of graphite from the volume of pencil we can get the volume of wood as well.
★ Sᴏʟᴜᴛɪᴏɴ :
Given Data
Diameter of pencil = 7 mm
As we know that radius is the half of diameter
So, radius (r) = 7/2 mm
Now, on the other hand
Diameter of Graphite = 1 mm
Radius (R) = 1/2 mm
Length of the pencil = 14 cm
[The length of the pencil is in cm. So we need to convert it into mm]
Hence, length = 140 mm
\begin{gathered} \star \quad \underline{ \boxed{ \green{\frak{Volume _{(graphite)} = \pi {r}^{2} h}}}} \\ \\ \dashrightarrow \frak{Volume _{(graphite)} = \frac{22}{7} \times {( \frac{1}{2}) }^{2} \times 140} \\ \\ \dashrightarrow \frak{Volume _{(graphite)} = \frac{22}{7} \times \frac{1}{4} \times 140} \\ \\ \qquad \frak{ \red{Volume _{(graphite)} = 110 \: m {m}^{3}}}\end{gathered}
⋆
Volume
(graphite)
=πr
2
h
⇢Volume
(graphite)
=
7
22
×(
2
1
)
2
×140
⇢Volume
(graphite)
=
7
22
×
4
1
×140
Volume
(graphite)
=110mm
3
Now, finding the volume of pencil
\begin{gathered} \star \quad \underline{ \boxed{ \green{\frak{Volume _{(pencil)} = \pi { \mathcal{R}}^{2} h}}}} \\ \\ \dashrightarrow \frak{Volume _{(pencil)} = \frac{22}{7} \times {( \frac{7}{2}) }^{2} \times 140} \\ \\ \dashrightarrow \frak{Volume _{(pencil)} = \frac{22}{7} \times \frac{49}{4} \times 140} \\ \\ \qquad \frak{ \pink{Volume _{(pencil)} = 5390\: m {m}^{3}}}\end{gathered}
⋆
Volume
(pencil)
=πR
2
h
⇢Volume
(pencil)
=
7
22
×(
2
7
)
2
×140
⇢Volume
(pencil)
=
7
22
×
4
49
×140
Volume
(pencil)
=5390mm
3
\frak{Volume _{(pencil)} = \frac{5390}{1000} = 5.39 \: c {m}^{3} }Volume
(pencil)
=
1000
5390
=5.39cm
3
Finding Volume of wood
\begin{gathered} \frak{Volume _{(wood)} = Volume _{(pencil)} - Volume _{(graphite)}} \\ \\ \frak{Volume _{(wood)} = 5390 \: m {m}^{3} - 110 \: m {m}^{3} } \\ \\ \blue{\frak{Volume _{(wood)} = 5280 \: m {m}^{3}}}\end{gathered}
Volume
(wood)
=Volume
(pencil)
−Volume
(graphite)
Volume
(wood)
=5390mm
3
−110mm
3
Volume
(wood)
=5280mm
3
Now,
\frak{Volume _{(wood)} = \frac{5280}{1000} = 5.28 \: c {m}^{3} }Volume
(wood)
=
1000
5280
=5.28cm
3
Required Answer is 5.28 cm³ and 5.39 cm³