All I know is that the answer is F. How do I get to the answer F?

There is a vendor at a carnival selling bags of cotton candy. Each bag of cotton candy has the same cost. She starts each day wi

umka21 [38]

Answer:

The cost of each cotton candy bag is $4.

Step-by-step explanation:

Consider the provided information.

Let, the vendor has $x$ dollar at the starting of each day.

Let, the cost of each cotton candy bag is $y$ .

Now when she sells a total of 12 bags, she has $128.

$x+12y=128$ .....(1)

After she sells a total of 20 bags, she has $160.

$x+20y=160$ ....(2)

Subtract equation 1 from equation 2.

$(x+20y)-(x+12y)=160-128$

$8y=32$

$y=4$

Hence, the cost of each cotton candy bag is $4.

6 0

3 years ago

A mattress store in Stafford purchased a mattress at a cost of $255 and marked it up 100%. Later on, Cassie bought the mattress.

skad [1K]

Answer: 278.18

sorry if it's wrong

Step-by-step explanation: if you divide 255 by 11 you get 23.18181818181

so if you simplify you get 23.18 so that is the tax plus the original payment which is 278.18

8 0

2 years ago

Read 2 more answers

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

2 years ago

If a| |b and c| |d, which pair of angles are congruent?

Wittaler [7]

Answer:

the ansewr is

Step-by-step explanation:

c

4 0

3 years ago

Read 2 more answers

What is the volume of a cone with a radius of 3 and a height of 11?

kondaur [170]

4 0

3 years ago