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3 years ago

What is the highest ranking for a graduating student?

SAT

1 answer:

Kamila [148]3 years ago

8 0

The valedictorian is often the student with the highest ranking grade point average or gpa for short.

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When passing, _____ only when you can see the tires of the vehicle you passed in your rearview mirror.

earnstyle [38]

Answer:

Explanation:

A. Return to your lane

3 0

3 years ago

Read 2 more answers

Fred bought 5 new baseball trading cards to add to his collection.the next day his dog ate half of his collection. there are now

Alika [10]

42 cards is after his dog ate half, so before that would be 42 * 2 = 84.

Even before that, he bought 5 new baseball cards, so add that to the 84 cards and you have 84 + 5 = 89.

Fred started out with 89 baseball cards.

4 0

4 years ago

Compare and contrast irish and chinese railroad workers.

Zielflug [23.3K]

Answer:

They speak different languages.

Explanation:

They live in different areas.

7 0

3 years ago

Given the equation representing a reaction: h + h h2 what occurs during this reaction?

tangare [24]

A bond is formed, and energy is released.

Explanation:

This is because energy is absorbed when a bond is broken. So, when these to Hydrogen combines a bond is formed and the energy is released,

Also, for a bond to be in broken energy, the energy is absorbed but when the bond is going to be formed, the energy is released instead.

8 0

2 years ago

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of len

Citrus2011 [14]

Answer:

The additional words is $2^n$

Explanation:

Given

$B(n) = 2^n$

Required

Determine the additional words; i.e. $B(n + 1) - B(n)$

From the given parameters, we have that;

B is a function of n

Such that;

$B(n) = 2^n$

To calculate $B(n+1)$ , we simply substitute n + 1 for n

$B(n) = 2^n$

$B(n + 1) = 2^{n + 1}$

Applying laws of indices

$B(n + 1) = 2^{n} * 2^1$

$B(n + 1) = 2^{n} * 2$

$B(n + 1) = 2(2^{n})$

Calculating Additional Binary Code;

$B(n + 1) - B(n)$

Substitute values for B(n + 1) and B(n)

$B(n + 1) - B(n) = 2(2^n) - 2^n$

Express $2^n$ as $2^ n * 1$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 1$

Express 1 as $2^0$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 2^0$

Factorize

$B(n + 1) - B(n) = 2^n(2 - 2^0)$

$B(n + 1) - B(n) = 2^n(2 - 1)$

$B(n + 1) - B(n) = 2^n(1)$

$B(n + 1) - B(n) = 2^n$

Hence, the additional words is $2^n$

8 0

3 years ago