The model represents an equation. What is the solution for the equation?

Surface area of solid cuboid 8cm 6cm 2cm

strojnjashka [21]

Answer:

152 $\text{cm}^2$

Step-by-step explanation:

We know that the total surface area of a cuboid is the sum of the surface area of all six of its faces.

We also know that each opposing side has the same surface area. Thus the total surface area is equal to:

$2*(8*6+8*2+6*2)\\\\=2*(48+16+12)\\=2*76\\=152$ ( $\text{cm}^2$ )

6 0

2 years ago

Which is part of the process in the formation of hail pellets? They are tossed up and down in clouds. They lose layers of ice an

kondaur [170]

Answer:

they are tossed up and down in clouds

3 0

2 years ago

McDonalds sells quarter-pound cheeseburgers for .99 cents last saturday.they sold 310 of these cheeseburgers. how many pounds of

Anvisha [2.4K]

The first thing you should do is calculate how many pounds of cheese a hamburger has
1 hamburger = 1/4 pound of cheese.
Then, you can make a rule of three to know how many pounds of cheese were sold:
1 ---> 1/4
310 ---> x
Clearing x we have:
x = ((310) / (1)) * (1/4) = 77.5
Answer
they sell last saturday 77.5 pounds of cheese

4 0

3 years ago

The cost of t-shirts varies directly with the number of t-shirts. The data is shown in the graph. If x = is number of t-shirts,

Lera25 [3.4K]

The answer is y = 5.5x, so B

3 0

3 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