Answer:
There’s only one way to get 12 when 2 dice are tossed, both have to equal 6. There are 6 ways tossing a single die can come out (1,2,3,4,5,6), so if you toss dice, the second die could have any one of six values with each of the numbers that could result from the first toss (e.g., 1 from die 1 and 1,2,3,4,5, or 6 from die 2). So, considering there are 6 ways to fill each of two slots, there are 6 x 6 = 36 possible outcomes of tossing two dice. Only one of them equals 12, so p(12 given 2 dice tossed) = 1/36 = 0.02777777777778.
For this sort of problem, you need to be familiar with prefixes.
So:
megabyte = 
bytes
kilobyte = 
bytes
and a regular byte would just be 
Now, you will need to do a conversion:
I'm going to explain this just in case, but we convert megabytes to regular bytes in the first half by using the information I gave you above. (In one megabyte, there are 
bytes!)
In the second fraction thought, remember one kilobyte is bytes. However, you usually only see one of each thing on the bottom of fractions, so you need to add a negative sign to the 3. (technically you could just do 
, but I think it is more correct to do write it out how I did).
*** They do equal the same thing though! Do whichever way is easier for you!
Now, your answer should be:
kilobytes (1000 kilobytes)
Hope this helped!
Answer:
300 cupcakes
Step-by-step explanation:
The value of this answer is .B
Answer:
∠1 ≅ ∠2 ⇒ proved down
Step-by-step explanation:
#12
In the given figure
∵ LJ // WK
∵ LP is a transversal
∵ ∠1 and ∠KWP are corresponding angles
∵ The corresponding angles are equal in measures
∴ m∠1 = m∠KWP
∴ ∠1 ≅ ∠KWP ⇒ (1)
∵ WK // AP
∵ WP is a transversal
∵ ∠KWP and ∠WPA are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠KWP = m∠WPA
∴ ∠KWP ≅ ∠WPA ⇒ (2)
→ From (1) and (2)
∵ ∠1 and ∠WPA are congruent to ∠KWP
∴ ∠1 and ∠WPA are congruent
∴ ∠1 ≅ ∠WPA ⇒ (3)
∵ WP // AG
∵ AP is a transversal
∵ ∠WPA and ∠2 are interior alternate angles
∵ The interior alternate angles are equal in measures
∴ m∠WPA = m∠2
∴ ∠WPA ≅ ∠2 ⇒ (4)
→ From (3) and (4)
∵ ∠1 and ∠2 are congruent to ∠WPA
∴ ∠1 and ∠2 are congruent
∴ ∠1 ≅ ∠2 ⇒ proved
