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5 nickels in a quarter
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How many different ways can you make change for a half dollar without using pennies?
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
What are the different combinations you can use to make 0.35 only using quarters dimes and nickels?
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
Why is there sometimes a remainder in division?
Because sometimes there will be things leftover and you can't split it all up in the question.
Can normal distribution be used if the data is not normal?
You can use a normal distribution to approximate a binomial distribution if conditions are met such as n*p and n*q is > or = to 5 & n >30.
If you have 4.35 in dimes nickels and quarters and there are 3 more dimes than nickels and 5 more quarters than nickels how many coins do you have?
5n + 10d + 25q = 435; d = n + 3 and q = n + 5 5n + 10(n + 3) + 25(n + 5) = 435 5n + 10n + 30 + 25n + 125 = 435 5n + 10n + 25n = 435 - 30 - 125 40n = 280 n = 7 so d = 10 and q = 12 (35c + $1.00 + $3.00 = $4.35)
What does 5 N in a Q mean?
Q(5n)
How many ways can you make change for a dollar using nickels dimes andor quarters?
There are 22 ways to make change from a dollar using nickels, dimes, and quarters. 1. 4 q 2. 10 d 3. 20 n 4. 2 q , 5 d 5. 3 q , 2 d , 1 n 6. 1 q , 7 d, 1 n 7. 9 d, 2 n 8. 8 d, 4 n 9. 7 d, 6 n 10. 6 d , 8 n 11. 5 d , 10 n 12. 4 d , 12 n 13. 2 d , 16 n 14. 1 d , 18 n 15. 5 n , 3 q 16. 3 n , 1 q , 6 d 17. 7 n , 1 q , 4 d 18. 9 n , 1 q , 3 d 19. 11 n , 1 q , 2 d 20. 13 n , 1 q , 1 d 21. 14n , 3 d 22. 15n , 1 q
How many different ways can you make change for a half dollar without using pennies?
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
A positive test charge of 5.00 E-5 C is placed in an electric field The force on it is 0.751 N The magnitude of the electric field at the location of the test charge is what?
We have to use F = E q Given F = 0.751 N, q = 5*10-5 C So required E = F /q = 0.751/5*10-5 Electric field at the location E = 15020 N C-1 or V m-1
What are the different combinations you can use to make 0.35 only using quarters dimes and nickels?
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
In a hypothesis for a proportion when may normality be assumed?
If n*p is greater than or equal to 5 AND n*q is greater than or equal to 5, you can use a normal distribution as an estimate for the binomial distribution. Recall, n is the number of trials, p is the probability of success of a trial, and q is 1-p.
James has 33 coins in his pocket all of them nickels and quarters If he has a total of 2.65 how many quarters does he have?
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
Why is there sometimes a remainder in division?
Because sometimes there will be things leftover and you can't split it all up in the question.
A letter that starts with N and that a Q in?
Nonequivalent is a word. It begins with N and contains the letter Q.
What are words that start with q and end with n?
Queen, quadrillion and qualification begin with the letter Q and end with the letter N. Question and quotation begin with Q and end with N.
Is the cube root of 5 irrational?
Yes it is. There is no fraction which, when cubed, equates to 5. Consider n = p/q where p and q are integers expressed in lowest terms (they are relatively prime). If n3 = 5, then p3 / q3 = 5. This equates to an integer if and only if q3 = 1 meaning q = 1 or if p3 is divisible by q3. The latter is impossible since p and q are relatively prime. Thus, for n to be the cube root of 5 and be rational, it must be an integer. No integer cubes to 5 (1, 8, ...). Thus, the cube root of 5 is irrational.
Can normal distribution be used if the data is not normal?
You can use a normal distribution to approximate a binomial distribution if conditions are met such as n*p and n*q is > or = to 5 & n >30.
