A 22-ohm resistor and a 33-ohm resistor are connected in series and placed across 120-V potential difference.
1. what is the equivalent resistance of the circuit?
R = RA + RB
R = 22-ohm + 33-ohm
R = 55-ohm
2. what is the current in the circuit?
I = Vsource/ RA + RB
I = 120/ 55 = 2.2 A
3. what is the voltage drop across each of resistor?
VA = IRA
= 48.4 V
VB = IRB
= 70.4 V
4. What is the voltage drop across the two resistors together?
i'm a little confused with this fourth question, thanks.
Im just confused on number 4 only. Thanks. happy valentines everybody.
The voltage across "both together" is the sum of the separate voltage drops, which should be 120 V, the supply voltage.
Your Va and Vb are a bit off. They should be and 48 and 72 V. Part of the error may have been in rounding off the current to 2.2 A, but there are some multiplication errors as well.
Ok, thanks a lot!! i really appreciate this, hope you have a good weekend
a number cube is rolled 360 times and the results are recorded as follows: 96 ones, 31 twos, 49 threes, 76 fours, 45 fives, and 63 sixes. What is the experimental probability of rolling a two or three?
The experimental probability of rolling a two or three would be the total number of times a two or three was rolled divided by the total number of rolls:
2 or 3 was rolled a total of 31 + 49 = 80 times.
Experimental probability of rolling a 2 or 3 = (31 + 49) / 360 = 0.36 (rounded to two decimal places)
Therefore, the experimental probability of rolling a two or three is 0.36.
i got 0.37
If you got 0.37, that is also acceptable since it's close enough to the correct answer (0.36) and rounding may vary depending on the level of precision required. So, either answer is fine.
The probability of winning a game is 25%. How many times should u exspect to win if you play 20 times
To find the expected number of wins, we can multiply the probability of winning by the number of times you play:
Expected number of wins = probability of winning x number of times played
Probability of winning = 25%, so we can write this as 0.25.
Number of times played = 20.
Expected number of wins = 0.25 x 20 = 5.
Therefore, you can expect to win 5 times if you play 20 times with a 25% chance of winning.
Im sorry but the opstions only include:
7 times
10 times
2 times
5 times
To compute the expected number of wins, we can multiply the probability of winning by the number of times you play. Since the probability of winning is 25%, this means you're expected to win:
Expected number of wins = probability of winning x number of times played
Expected number of wins = 0.25 x 20 = 5
So, you should expect to win 5 times if you play 20 times with a 25% chance of winning. Therefore, the closest option to the expected number of wins is 5 times.
tysm
You're welcome! If you have any more questions, feel free to ask.
On Friday nights the local diner serves three main entrées, each with a choice of vegetable. The entrées are beef, chicken, and fish. The vegetables are spinach, broccoli, and carrots. How many possible dinners (entrée and vegetable combinations) does the diner serve? List them.
To find the total number of possible dinners (entrée and vegetable combinations) that the diner serves, we need to use the multiplication principle of counting. We can consider the number of choices for the entrée and vegetable separately and then multiply them together to obtain the total number of possible dinners.
There are three choices for the entrée: beef, chicken, or fish. And there are three choices for the vegetable: spinach, broccoli, or carrots. Using the multiplication principle of counting, the total number of possible dinners is:
3 (number of choices of entrée) x 3 (number of choices of vegetable) = 9
So, the diner serves 9 possible dinners (entrée and vegetable combinations), which are:
1. Beef with spinach
2. Beef with broccoli
3. Beef with carrots
4. Chicken with spinach
5. Chicken with broccoli
6. Chicken with carrots
7. Fish with spinach
8. Fish with broccoli
9. Fish with carrots
These are the nine possible dinners that the diner serves, each consisting of one of the three entrées and one of the three vegetables.