Johnny and Dawn were in a Math competition.

The radius of a nitrogen atom is 5.8*10^-11 meters, and the radius of a beryllium atom is 1.12*10^-10 meters. Which atom has a l

Burka [1]

Nitrogen Radius = 5.8 x 10⁻¹¹ m

Beryllium Radius = 1.12 x 10⁻¹⁰ m

Let's find the quotient of N/Be :

(5.8x10⁻¹¹)/(1.12x10⁻¹⁰). But 10⁻¹¹/10⁻¹⁰ = 10⁽⁻¹¹⁺¹⁰⁾ = 10⁻¹ = 1/10 = 0.1

→ (5.8/1.12).(0.1) = 0.58/1.12 = 0.518.

Conclusion: the radius of Be is almost double than the radius of N

5 0

3 years ago

How many variables are in this equation: 2y + 5 = 5n

pickupchik [31]

Answer:

There are 2 variables in the equation. The variables are y and n

5 0

3 years ago

Read 2 more answers

Given a standard deck of 52 cards, 3 cards are dealt without replacement. Using this situation, answer the questions below.<b

kherson [118]

Given that 3 cards are dealt without replacement in a standard deck of 52 cards.

Part A:

There are 4 queens in a standard deck of 52 card, thus the probability that the first card is a queen is given by 4 / 52 = 1 / 13.

Since, the first card is not replaced, thus there are 3 queens remaining and 51 ards remaining in total, thus the probability that the second card is a queen is given by 3 / 51 = 1 / 17

Similarly the probability that the third card is a queen is given by 2 / 50 = 1 / 25.

Therefore, the probability that all three cards are queens is given by

$\frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} = \frac{1}{5525}$

Part B:

Yes the probability of drawing a queen of heart is independent of the probability of drawing a queen of diamonds because they are separate cards and drawing one of the cards does not in any way affect the chance of drawing the other card.

Part C:

Given that the first card is a queen, then there are 3 queens remaining out of 51 cards remaining, thus the number of cards that are not queen is 51 - 3 = 48 cards.

Therefore, if the first card is a queen, the probability that the second card will not be a queen is given by 48 / 51 = 16 / 17

Part D:

Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining.

Therefore, if two of the three cards are queens ,the probability that you will be dealt three queens is given by 2 / 50 = 1 / 25 = 0.04

Part E:

Given that the first two card are queens, then there are 2 queens remaining out of 50 cards remaining, thus the number of cards that are not queen is 50 - 2 = 48 cards.

Therefore, if two of the three cards are queens ,the probability that the other card is not a queen is given by 48 / 50 = 24 / 25 = 0.96

8 0

3 years ago

What is the slope of the line?

serg [7]

Answer:

i hated iready xD the answer should be 2

Step-by-step explanation:

if you use ride over run the rise would be 2 and the run is 1 $\frac{2}{1}$ = 2