The total cost of a shirt and an $8 pair of socks

Jake made a wooden cone with a diameter of 6 inches and a height of 4.5 inches. What is the volume of this cone?

serious [3.7K]

Answer:

42.42in³

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, a cone

Step one

We know that the volume of a cone is

V= 1/3πr²h

Given data

Diameter d= 6 inches

Radius r= diameter/2= 6/2= 3 inches

Height h = 4.5 inches

Step two

Substituting our given data we have

Volume = 1/3*3.142*3²*4.5

V= 1/3*3.142*9*4.5

V= 127.251/3

V= 42.42in³

7 0

3 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

Answer:

Question 1: $P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}$

Question 2:

A. $P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

B. $P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

C. $P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

Step-by-step explanation:

Conditional probability is defined by

$P(A|B)= \frac{P(A and B)}{P(B)}$

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

then

$P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

so

$P((YorZ) and B)= \frac{3}{16}$

By definition,

$P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}$

3 0

3 years ago

Please help me im stuck on this

Strike441 [17]

Answer:

Step-by-step explanation:

dddfykb

7 0

1 year ago

(x +y) + Z commutative property of addition

goldenfox [79]

Answer:

Step-by-step explanation:

commutative property of addition : x + y = y + x

example : 2+3 =5 and 3+2 =5 then 2+3=3+2

associative property of addition

x + y + z = (x + y) + z = x + (y + z)

example: 2+3+4 = (2+3)+4 = 5 + 4 =9

2+3+4 = 2+(3+4) = 2 + 7 = 9

8 0

3 years ago

The area of a rectangle is 42 square inches and one side is 12 inches long. Find the perimeter of the rectangle. Show your work.

LenaWriter [7]

I might not be correct but i think Perimeter=31. Since the problem tells me that the area is 42 square inches i divided the given area (42 inches) by the given side length (12 inches), and I got 3.5. I did this because the formula for area is a=lw therefore 12×3.5=42 inches. Now that I know the missing measurement of the other side length, I can plug the numbers I have into either one of these equations.
P=SIDE++SIDE+SIDE+SIDE OR P=2(length)+ 2(width). Please let me know whether or not you understand or have any other questions I will be more than happy to answer. Later when you reach 9th grade and become a freshmen you will learn theorems that will help make it a little bit easier and they will also remove A LOT of the algebraic work that's required to solve questions such as this one. My name is Takira and I go to LaGuardia High School for Performing Arts. Please let me know if you have any more questions.

7 0

3 years ago