coin - A unit of currency

$25.00 is how much in quarters

$25.00 is how much in quarters
We type [URL='https://www.mathcelebrity.com/coincon.php?quant=25&type=dollar&pl=Calculate']25 dollars in our math engine[/URL] and we get:
[B]100 quarters[/B]

$3.75 in quarters and nickles in her car. The number of nickles is fifteen more than the number of q

$3.75 in quarters and nickels in her car. The number of nickels is fifteen more than the number of quarters. How many of each type of coin does she have?
Let the number of nickels be n, and the number of quarters be q. We know nickels are 0.05, and quarters are 0.25. We're given:
[LIST=1]
[*]n = q + 15
[*]0.05n + 0.25q = 3.75
[/LIST]
Substituting (1) into (2), we get:
0.05(q + 15) + 0.25q = 3.75
0.05q + 0.75 + 0.25q = 3.75
Combine like term:
0.3q + 0.75 = 3.75
[URL='https://www.mathcelebrity.com/1unk.php?num=0.3q%2B0.75%3D3.75&pl=Solve']Typing this equation into our calculator[/URL], we get:
[B]q = 10[/B]
Substituting q = 10 into Equation (1), we get:
n = 10 + 15
[B]n = 25[/B]

2 coins are tossed. Find the probability of getting 1 head and 1 tail

2 coins are tossed. Find the probability of getting 1 head and 1 tail
We can either flip HT or TH. Let's review probabilities:
[LIST]
[*]HT = 1/2 * 1/2 = 1/4 <-- We multiply since each event is independent
[*]TH = 1/2 * 1/2 = 1/4 <-- We multiply since each event is independent
[/LIST]
P(1 H, 1 T) = P(HT) + P(TH)
P(1 H, 1 T) = 1/4 + 1/4
P(1 H, 1 T) = 2/4
[URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F4&frac2=3%2F8&pl=Simplify']Using our fraction simplifier[/URL], we can reduce 2/4 to 1/2
P(1 H, 1 T) = [B]1/2[/B]

2 times as many dimes as quarters and they have a combined value of 180 cents, how many of each coin

2 times as many dimes as quarters and they have a combined value of 180 cents, how many of each coin does he have?
Let d be the number of dimes. Let q be the number of quarters. We're given two equations:
[LIST=1]
[*]d = 2q
[*]0.1d + 0.25q = 180
[/LIST]
Substitute (1) into (2):
0.1(2q) + 0.25q = 180
0.2q + 0.25q = 180
[URL='https://www.mathcelebrity.com/1unk.php?num=0.2q%2B0.25q%3D180&pl=Solve']Typing this equation into the search engine[/URL], we get:
[B]q = 400[/B]
Now substitute q = 400 into equation 1:
d = 2(400)
[B]d = 800[/B]

79 cents in 7 coins

3 quarters = 0.25 * 3 = 0.75
4 pennies = 0.01 * 4 = 0.04
Total Coin Value = 0.75 + 0.04 = 0.79
[B]3 quarters, 4 pennies[/B]

A bag of quarters and nickels is worth $8.30. There are two less than three times as many quarters a

A bag of quarters and nickels is worth $8.30. There are two less than three times as many quarters as nickels. How many of the coins must be quarters?
Assumptions and givens:
[LIST]
[*]Let the number of quarters be q
[*]Let the number of nickels be n
[/LIST]
We have two equations:
[LIST=1]
[*]0.05n + 0.25q = 8.30
[*]n = 3q - 2 [I](Two less than Three times)[/I]
[/LIST]
Plug in equation (2) into equation (1) for q to solve this system of equations:
0.05(3q - 2) + 0.25q = 8.30
To solve this equation for q, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.05%283q-2%29%2B0.25q%3D8.30&pl=Solve']type it in our search engine[/URL] and we get:
q = [B]21[/B]

A coin is tossed 3 times. a. Draw a tree diagram and list the sample space that shows all the possib

A coin is tossed 3 times. a. Draw a tree diagram and list the sample space that shows all the possible outcomes
[URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']type in "toss a coin 3 times" and pick the probability option[/URL].

A coin is tossed and a die is rolled. Find the probability pf getting a head and a number greater th

A coin is tossed and a die is rolled. Find the probability pf getting a head and a number greater than 4.
Since each event is independent, we multiply the probabilities of each event.
P(H) = 0.5 or 1/2
P(Dice > 4) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3
P(H) AND P(Dice > 4) = 1/2 * 1/3 = [B]1/6[/B]

A collection of nickels and dime has a total value of $8.50. How many coins are there if there are 3

A collection of nickels and dime has a total value of $8.50. How many coins are there if there are 3 times as many nickels as dimes.
Let n be the number of nickels. Let d be the number of dimes. We're give two equations:
[LIST=1]
[*]n = 3d
[*]0.1d + 0.05n = 8.50
[/LIST]
Plug equation (1) into equation (2) for n:
0.1d + 0.05(3d) = 8.50
Multiply through:
0.1d + 0.15d = 8.50
[URL='https://www.mathcelebrity.com/1unk.php?num=0.1d%2B0.15d%3D8.50&pl=Solve']Type this equation into our search engine[/URL] and we get:
[B]d = 34[/B]
Now, we take d = 34, and plug it back into equation (1) to solve for n:
n = 3(34)
[B]n = 102[/B]

A crate contains 300 coins and stamps. The coins cost $3 each and the stamps cost $1.5 each. The tot

A crate contains 300 coins and stamps. The coins cost $3 each and the stamps cost $1.5 each. The total value of the items is $825. How many coins are there?
Let c be the number of coins, and s be the number of stamps. We're given:
[LIST=1]
[*]c + s = 300
[*]3c + 1.5s = 825
[/LIST]
We have a set of simultaneous equations, or a system of equations. We can solve this 3 ways:
[LIST=1]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Substitution']Substitution Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Elimination']Elimination Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Cramers+Method']Cramers Method[/URL]
[/LIST]
No matter which way we pick, we get:
s = 50
c = [B]250[/B]

A die and a coin are tossed. What is the probability of getting a 6 and a tail?

A die and a coin are tossed. What is the probability of getting a 6 and a tail?
Roll a 6:
P(6) = 1/6
Flip a tail:
P(T) = 1/2
Probability of getting a 6 and a tail:
Since both events are independent, we have:
P(6 and T) = P(6) * P(T)
P(6 and T) = 1/6 * 1/2
P(6 and T) = [B]1/12[/B]

A fair coin is tossed 4 times. a) How many outcomes are there in the sample space? b) What is the pr

A fair coin is tossed 4 times.
a) How many outcomes are there in the sample space?
b) What is the probability that the third toss is heads, given that the first toss is heads?
c) Let A be the event that the first toss is heads, and B be the event that the third toss is heads. Are A
and B independent? Why or why not?
a) 2^4 = [B]16[/B] on our [URL='http://www.mathcelebrity.comcointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=+4&calc=5&montect=+500&pl=Calculate+Probability']coin toss calculator[/URL]
b) On the link above, 4 of those outcomes have H and H in toss 1 and 3. So it's [B]1/4 or 0.25[/B]
c) [B]Yes, each toss is independent of each other.[/B]

A fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a h

A fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a head?
Since you always flip a head, we have:
P(Head) = [B]1 or 100%[/B]

A jar contains 80 nickels and dimes worth $6.40. How many of each kind of coin are in the jar?

A jar contains 80 nickels and dimes worth $6.40. How many of each kind of coin are in the jar?
Using our [URL='http://www.mathcelebrity.com/coin-word-problem.php?coinvalue=6.40&cointot=80&coin1=nickels&coin2=dimes&pl=Calculate+Coin+Quantities']coin combination word problem calculator[/URL], we get:
[LIST]
[*][B]48 dimes[/B]
[*][B]32 nickels[/B]
[/LIST]

A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the pro

A number cube is rolled and a coin is tossed. The number cube and the coin are fair. What is the probability that the number rolled is greater than 3 and the coin toss is heads? Write your answer as a fraction in simplest form
Let's review the vitals of this question:
[LIST]
[*]The probability of heads on a fair coin is 1/2.
[*]On a fair die, greater than 3 means either 4, 5, or 6. Any die roll face is a 1/6 probability.
[*]So we have a combination of outcomes below:
[/LIST]
Outcomes
[LIST=1]
[*]Heads and 4
[*]Heads and 5
[*]Heads and 6
[/LIST]
For each of the outcomes, we assign a probability. Since the coin flip and die roll are independent, we multiply the probabilities:
[LIST=1]
[*]P(Heads and 4) = 1/2 * 1/6 = 1/12
[*]P(Heads and 5) = 1/2 * 1/6 = 1/12
[*]P(Heads and 6) = 1/2 * 1/6 = 1/12
[/LIST]
Since we want any of those events, we add all three probabilities
1/12 + 1/12 + 1/12 = 3/12
This fraction is not simplified. S[URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F12&frac2=3%2F8&pl=Simplify']o we type this fraction into our search engine, and choose Simplify[/URL].
We get a probability of [B]1/4[/B].
By the way, if you need a decimal answer or percentage answer instead of a fraction, we type in the following phrase into our search engine:
[URL='https://www.mathcelebrity.com/perc.php?num=1&den=4&pcheck=1&num1=+16&pct1=+80&pct2=+35&den1=+90&pct=+82&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']1/4 to decimal[/URL]
Alternative Answers:
[LIST]
[*]For a decimal, we get [B]0.25[/B]
[*]For a percentage, we get [B]25%[/B]
[/LIST]

A parking meter contains 27.05 in quarters and dimes. All together there are 146 coins. How many of

A parking meter contains 27.05 in quarters and dimes. All together there are 146 coins. How many of each coin are there?
Let d = the number of dimes and q = the number of quarters. We have two equations:
(1) d + q = 146
(2) 0.1d + 0.25q = 27.05
Rearrange (1) into (3) solving for d
(3) d = 146 - q
Substitute (3) into (2)
0.1(146 - q) + 0.25q = 27.05
14.6 - 0.1q + 0.25q = 27.05
Combine q's
0.15q + 14.6 = 27.05
Subtract 14.6 from each side
0.15q = 12.45
Divide each side by 0.15
[B]q = 83[/B]
Plugging that into (3), we have:
d = 146 - 83
[B]d = 63[/B]

A pile of coins, consisting of quarters and half dollars, is worth 11.75. If there are 2 more quarte

A pile of coins, consisting of quarters and half dollars, is worth 11.75. If there are 2 more quarters than half dollars, how many of each are there?
Let h be the number of half-dollars and q be the number of quarters. Set up two equations:
(1) q = h + 2
(2) 0.25q + 0.5h = 11.75
[U]Substitute (1) into (2)[/U]
0.25(h + 2) + 0.5h = 11.75
0.25h + 0.5 + 0.5h = 11.75
[U]Group h terms[/U]
0.75h + 0.5 = 11.75
[U]Subtract 0.5 from each side[/U]
0.75h = 11.25
[U]Divide each side by h[/U]
[B]h = 15[/B]
[U]Substitute h = 15 into (1)[/U]
q = 15 + 2
[B]q = 17[/B]

A suitcase contains nickels, dimes and quarters. There are 2&1/2 times as many dimes as nickels and

A suitcase contains nickels, dimes and quarters. There are 2&1/2 times as many dimes as nickels and 5 times the number of quarters as the number of nickels. If the coins have a value of $24.80, how many nickels are there in the suitcase?
Setup number of coins:
[LIST]
[*]Number of nickels = n
[*]Number of dimes = 2.5n
[*]Number of quarters = 5n
[/LIST]
Setup value of coins:
[LIST]
[*]Value of nickels = 0.05n
[*]Value of dimes = 2.5 * 0.1n = 0.25n
[*]Value of quarters = 5 * 0.25n = 1.25n
[/LIST]
Add them up:
0.05n + 0.25n + 1.25n = 24.80
Solve for [I]n[/I] in the equation 0.05n + 0.25n + 1.25n = 24.80
[SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE]
(0.05 + 0.25 + 1.25)n = 1.55n
[SIZE=5][B]Step 2: Form modified equation[/B][/SIZE]
1.55n = + 24.8
[SIZE=5][B]Step 3: Divide each side of the equation by 1.55[/B][/SIZE]
1.55n/1.55 = 24.80/1.55
n = [B]16[/B]
[B]
[URL='https://www.mathcelebrity.com/1unk.php?num=0.05n%2B0.25n%2B1.25n%3D24.80&pl=Solve']Source[/URL][/B]

Ahmad has a jar containing only 5-cent and 20-cent coins. In total there are 31 coins with a total v

Ahmad has a jar containing only 5-cent and 20-cent coins. In total there are 31 coins with a total value of $3.50. How many of each type of coin does Ahmad have?
Let the number of 5-cent coins be f.
Let the number of 20-cent coins be t.
We're given two equations:
[LIST=1]
[*]f + t = 31
[*]0.05f + 0.2t = 3.50
[/LIST]
We can solve this simultaneous system of equations 3 ways:
[LIST]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Substitution']Substitution Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Elimination']Elimination Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Cramers+Method']Cramers Rule[/URL]
[/LIST]
No matter which method we choose, we get:
[LIST]
[*][B]f = 18[/B]
[*][B]t = 13[/B]
[/LIST]

Anna has 50 coins in her piggy bank. She notices that she only has dimes and pennies. If she has exa

Anna has 50 coins in her piggy bank. She notices that she only has dimes and pennies. If she has exactly four times as many pennies as dimes, how many pennies are in her piggy bank?
Let d be the number of dimes, and p be the number of pennies. We're given:
[LIST=1]
[*]d + p = 50
[*]p = 4d
[/LIST]
Substitute (2) into (1)
d + 4d = 50
[URL='https://www.mathcelebrity.com/1unk.php?num=d%2B4d%3D50&pl=Solve']Type that equation into our search engine[/URL]. We get:
d = 10
Now substitute this into Equation (2):
p = 4(10)
[B]p = 40[/B]

Antonio has a change jar that contains $3.65 in half dollars and nickels. He has 7 more nickels than

Antonio has a change jar that contains $3.65 in half dollars and nickels. He has 7 more nickels than half dollars. How many of each type of coin does he have?
Let h be half dollars
Let n be nickels
We're given two equations:
[LIST=1]
[*]n = h + 7
[*]0.5h + 0.05n = 3.65
[/LIST]
Substitute equation (1) into equation (2) for n:
0.5h + 0.05(h + 7) = 3.65
To solve this equation for h, we[URL='https://www.mathcelebrity.com/1unk.php?num=0.5h%2B0.05%28h%2B7%29%3D3.65&pl=Solve'] type it in our search engine[/URL] and we get:
h = [B]6
[/B]
To get n, we substitute h = 6 into equation (1) above:
n = 6 + 7
n = [B]13[/B]

At the movie theater, Celeste bought 2 large drinks and 2 large popcorns for $8.50. She paid with a

At the movie theater, Celeste bought 2 large drinks and 2 large popcorns for $8.50. She paid with a twenty-dollar bill. What is the fewest number of bills and coins that she could have received as change?r of bills and coins that she could have received as change?
Calculate change:
Change = Amount Paid - Bill
Change = $20.00 - $8.50
Change = $11.50
Largest bill we can start with is a 10 dollar bill:
$11.50 - 10 = $1.50
Next largest bill is a $1 bill
$1.50 - $1 = 0.50
Now we're down to coins. Largest coin(s) we can use are quarters (assuming no half-dollars)
2 quarters equals 0.50
0.50 - 0.50 = 0
[U]Therefore, our answer is:[/U]
[B]Ten dollar Bill, 1 dollar bill, and 2 quarters[/B]

Bob has half as many quarters as dimes. He has $3.60. How many of each coin does he have?

Bob has half as many quarters as dimes. He has $3.60. How many of each coin does he have?
Let q be the number of quarters. Let d be the number of dimes. We're given:
[LIST=1]
[*]q = 0.5d
[*]0.25q + 0.10d = 3.60
[/LIST]
Substitute (1) into (2):
0.25(0.5d) + 0.10d = 3.60
0.125d + 0.1d = 3.6
Combine like terms:
0.225d = 3.6
[URL='https://www.mathcelebrity.com/1unk.php?num=0.225d%3D3.6&pl=Solve']Typing this equation into our search engine[/URL], we're given:
[B]d = 16[/B]
Substitute d = 16 into Equation (1):
q = 0.5(16)
[B]q = 8[/B]

Change Counting

Free Change Counting Calculator - This shows you how to make change using the least amount of bills/coins by taking a bill amount and a cash tendered amount from a customer and figuring out the fastest way to make change. Maximum denomination is $100

Charlie collects rare coins. Of his display cases, 3 cases contain 15 coins each and 4 cases contain

Charlie collects rare coins. Of his display cases, 3 cases contain 15 coins each and 4 cases contain 17 coins each. On Monday, Charlie bought 10 new coins for his collection. How many total coins does he have in his collection now?
We have cases * coins + extra coins:
3(15) + 4(17) + 10
45 + 68 + 10
[B]123 coins[/B]

Coin Amount to Denomination

Free Coin Amount to Denomination Calculator - Takes a money value and using the highest possible bills/coins, constructs the amount using bills and coins.

Coin Combinations

Free Coin Combinations Calculator - Given a selection of coins and an amount, this determines the least amount of coins needed to reach that total.

Coin Denomination Conversions

Free Coin Denomination Conversions Calculator - This caculator converts between the following coin denominations:

* Penny

* Nickel

* Dime

* Quarter

* Half-Dollar

* Dollar

* Penny

* Nickel

* Dime

* Quarter

* Half-Dollar

* Dollar

Coin Toss Probability

Free Coin Toss Probability Calculator - This calculator determines the following coin toss probability scenarios

* Coin Toss Sequence such as HTHHT

* Probability of x heads and y tails

* Probability of at least x heads in y coin tosses

* Probability of at least x tails in y coin tosses

* Probability of no more than x heads in y coin tosses

* Probability of no more than x tails in y coin tosses

* (n) Coin Tosses with a list of scenario results displayed

* Monte Carlo coin toss simulation

* Coin Toss Sequence such as HTHHT

* Probability of x heads and y tails

* Probability of at least x heads in y coin tosses

* Probability of at least x tails in y coin tosses

* Probability of no more than x heads in y coin tosses

* Probability of no more than x tails in y coin tosses

* (n) Coin Tosses with a list of scenario results displayed

* Monte Carlo coin toss simulation

Coin Total Word Problems

Free Coin Total Word Problems Calculator - This word problem lesson solves for a quantity of two coins totaling a certain value with a certain amount more or less of one coin than another

Coin Values

Free Coin Values Calculator - This calculates the total value of a given amount of:

* Pennies

* Nickels

* Dimes

* Quarters

* Half-Dollars

* Dollars

* Pennies

* Nickels

* Dimes

* Quarters

* Half-Dollars

* Dollars

Coin Word Problems

Free Coin Word Problems Calculator - This word problem lesson solves for a quantity of two coins totaling a certain value

Erin has 72 stamps in her stamp drawer along with a quarter, two dimes and seven pennies. She has 3

Erin has 72 stamps in her stamp drawer along with a quarter, two dimes and seven pennies. She has 3 times as many 3-cent stamps as 37-cent stamps and half the number of 5-cent stamps as 37-cent stamps. The value of the stamps and coins is $8.28. How many 37-cent stamps does Erin have?
Number of stamps:
[LIST]
[*]Number of 37 cent stamps = s
[*]Number of 3-cent stamps = 3s
[*]Number of 5-cent stamps = 0.5s
[/LIST]
Value of stamps and coins:
[LIST]
[*]37 cent stamps = 0.37s
[*]3-cent stamps = 3 * 0.03 = 0.09s
[*]5-cent stamps = 0.5 * 0.05s = 0.025s
[*]Quarter, 2 dime, 7 pennies = 0.52
[/LIST]
Add them up:
0.37s + 0.09s + 0.025s + 0.52 = 8.28
Solve for [I]s[/I] in the equation 0.37s + 0.09s + 0.025s + 0.52 = 8.28
[SIZE=5][B]Step 1: Group the s terms on the left hand side:[/B][/SIZE]
(0.37 + 0.09 + 0.025)s = 0.485s
[SIZE=5][B]Step 2: Form modified equation[/B][/SIZE]
0.485s + 0.52 = + 8.28
[SIZE=5][B]Step 3: Group constants:[/B][/SIZE]
We need to group our constants 0.52 and 8.28. To do that, we subtract 0.52 from both sides
0.485s + 0.52 - 0.52 = 8.28 - 0.52
[SIZE=5][B]Step 4: Cancel 0.52 on the left side:[/B][/SIZE]
0.485s = 7.76
[SIZE=5][B]Step 5: Divide each side of the equation by 0.485[/B][/SIZE]
0.485s/0.485 = 7.76/0.485
s = [B]16[/B]
[URL='https://www.mathcelebrity.com/1unk.php?num=0.37s%2B0.09s%2B0.025s%2B0.52%3D8.28&pl=Solve']Source[/URL]

Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head?

Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head? Give your answer in its simplest form.
Probability of a 5 is 1/6
Probability of a head is 1/2
Since each event is independent, we get the total probability by multiplying both together:
P(5,H) = 1/6 * 1/2
P(5,H) = [B]1/12[/B]

flip 7 coins How many total outcomes are there

flip 7 coins How many total outcomes are there
A flip of a coin has 2 outcomes, heads or tails. Since each outcome is independent of the other outcomes, we multiply each flip by 2 outcomes:
Total outcomes = 2 * 2 * 2 * 2 * 2 * 2 * 2
Total outcomes = 2^7
Total outcomes = [B]128[/B]

Four coins are flipped. What is the probability of the coins all landing on heads

Four coins are flipped. What is the probability of the coins all landing on heads
The probability of one head is 1/2. Since all 4 flips are independent, we multiply each flip probability:
P(HHHH) = 1/2 * 1/2 * 1/2 * 1/2
P(HHHH) = [B]1/16[/B]

Gayle has 36 coins, all nickels and dimes, worth $2.40. How many dimes does she have?

Gayle has 36 coins, all nickels and dimes, worth $2.40. How many dimes does she have?
Set up our given equations using n as the number of nickels and d as the number of dimes:
[LIST=1]
[*]n + d = 36
[*]0.05n + 0.1d = 2.40
[/LIST]
Use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+d+%3D+36&term2=0.05n+%2B+0.1d+%3D+2.40&pl=Cramers+Method']simultaneous equations calculator[/URL] to get:
n = 24
[B]d = 12[/B]

How many dimes must be added to a bag of 200 nickels so that the average value of the coins in the b

How many dimes must be added to a bag of 200 nickels so that the average value of the coins in the bag is 8 cents?
200 nickels has a value of 200 * 0.05 = $10.
Average value of coins is $10/200 = 0.05
Set up our average equation, where we have total value divided by total coins:
(200 * 0.05 + 0.1d)/(200 + d) = 0.08
Cross multiply:
16 + 0.08d = 10 + 0.1d
Using our [URL='http://www.mathcelebrity.com/1unk.php?num=16%2B0.08d%3D10%2B0.1d&pl=Solve']equation solver[/URL], we get:
[B]d = 300[/B]

How many nickels are in 3 quarters and 2 dimes

How many nickels are in 3 quarters and 2 dimes
[URL='https://www.mathcelebrity.com/coinvalue.php?p=&n=&d=2&q=3&h=&dol=&pl=Calculate+Coin+Value']3 quarters and 2 dimes[/URL] = 0.95
Since a nickel is 0.05, we have:
Number of nickels = 0.95/0.05
Number of nickels = [B]19[/B]

how many nickels are there in 10 dimes

how many nickels are there in 10 dimes
Using our[URL='https://www.mathcelebrity.com/coincon.php?quant=10&type=dime&pl=Calculate'] coin conversions calculator[/URL], we see that:
10 dimes = [B]20 nickels[/B]

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If 3 coins are flipped simultaneously, the probability of having three tails is

If 3 coins are flipped simultaneously, the probability of having three tails is...
The probability of flipping a head is 1/2. Since each coin flip is independent, we multiply the probabilities together of the three coin flips:
P(HHH) = 1/2 * 1/2 * 1/2
P(HHH) = [B]1/8[/B]

If a jar of coins contains 50 half-dollars and 120 quarters, what is the monetary value of the coins

If a jar of coins contains 50 half-dollars and 120 quarters, what is the monetary value of the coins?
We use our [URL='https://www.mathcelebrity.com/coinvalue.php?p=+&n=+&d=+&q=120&h=+50&dol=+&pl=Calculate+Coin+Value']coin values calculator[/URL], and we get:
[B]$55.00[/B]

if flip a coin 4 times, what is the probability of getting all 4 tails

if flip a coin 4 times, what is the probability of getting all 4 tails.
P(Tails) = 1/2
Each flip is independent, so we have:
[URL='https://www.mathcelebrity.com/cointoss.php?hts=TTTT&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']P(TTTT)[/URL] = [B]1/16[/B]

If Quinn has 4 times as many quarters as nickels and they have a combined value of 525 cents, how ma

If Quinn has 4 times as many quarters as nickels and they have a combined value of 525 cents, how many of each coin does he have?
Using q for quarters and n for nickels, and using 525 cents as $5.25, we're given two equations:
[LIST=1]
[*]q = 4n
[*]0.25q + 0.05n = 5.25
[/LIST]
Substitute equation (1) into equation (2) for q:
0.25(4n) + 0.05n = 5.25
Multiply through and simplify:
n + 0.05n = 5.25
To solve this equation for n, we [URL='https://www.mathcelebrity.com/1unk.php?num=n%2B0.05n%3D5.25&pl=Solve']type it in our search engine[/URL] and we get:
n = [B]5
[/B]
To get q, we plug in n = 5 into equation (1) above:
q = 4(5)
q = [B]20[/B]

If two coins are flipped, what is the probability that there will not be two heads?

If two coins are flipped, what is the probability that there will not be two heads?
There's only one way to flip 2 coins and get 2 heads:
P(HH) = 1/2 * 1/2 = 1/4
Which means the probability of NOT getting 2 heads is:
1 - 1/4 = [B]3/4[/B]

If you toss a fair coin 6 times, what is the probability of getting all tails?

If you toss a fair coin 6 times, what is the probability of getting all tails?
We [URL='https://www.mathcelebrity.com/cointoss.php?hts=TTTTTT&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']type in our search engine [I]TTTTTT [/I]and we get[/URL]:
P(TTTTTT) = [B]1/64 or 0.015625[/B]

Jack has 34 bills and coins in 5’s and 2’s. The total value is $116. How many 5 dollar bills does he

Jack has 34 bills and coins in 5’s and 2’s. The total value is $116. How many 5 dollar bills does he have?
Let the number of 5 dollar bills be f. Let the number of 2 dollar bills be t. We're given two equations:
[LIST=1]
[*]f + t = 34
[*]5f + 2t = 116
[/LIST]
We have a system of equations, which we can solve 3 ways:
[LIST=1]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Substitution']Substitution Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Elimination']Elimination Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Cramers+Method']Cramer's Rule[/URL]
[/LIST]
No matter which method we choose, we get the same answers:
[LIST]
[*][B]f = 16[/B]
[*][B]t = 18[/B]
[/LIST]

Juan has d dimes and q quarters in his pocket. The total value of the coins is less than $14.75. Whi

Juan has d dimes and q quarters in his pocket. The total value of the coins is less than $14.75. Which inequality models this situation?
[U]Let d be the number of dimes and q be the number of quarters[/U]
[B]0.1d + 0.25q < 14.75[/B]

Juan has d dimes and q quarters in his pocket. The total value of the coins is less than $14.75. Whi

Juan has d dimes and q quarters in his pocket. The total value of the coins is less than $14.75. Which inequality models this situation?
Since dimes are worth $0.10 and quarters are worth $0.25, we have:
[B]0.10d + 0.25q < 14.75[/B]

Julio had a coin box that consisted of only quarters and dimes. The number of quarters was three tim

Julio had a coin box that consisted of only quarters and dimes. The number of quarters was three times the number of dimes. If the number of dimes is n, what is the value of coins in the coin box?
Set up monetary value:
[LIST]
[*]Value of the dimes = 0.1n
[*]Value of the quarters = 0.25 * 3n = 0.75n
[/LIST]
Add them together
[B]0.85n[/B]

Kendra has $5.70 in quarters and nickels. If she has 12 more quarters than nickels, how many of each

Kendra has $5.70 in quarters and nickels. If she has 12 more quarters than nickels, how many of each coin does she have?
Let n be the number of nickels and q be the number of quarters. We have:
[LIST=1]
[*]q = n + 12
[*]0.05n + 0.25q = 5.70
[/LIST]
Substitute (1) into (2)
0.05n + 0.25(n + 12) = 5.70
0.05n + 0.25n + 3 = 5.70
Combine like terms:
0.3n + 3 = 5.70
Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.3n%2B3%3D5.70&pl=Solve']equation calculator[/URL], we get [B]n = 9[/B].
Substituting that back into (1), we get:
q = 9 + 12
[B]q = 21[/B]

Kevin and randy have a jar containing 41 coins, all of which are either quarters or nickels. The tot

Kevin and randy have a jar containing 41 coins, all of which are either quarters or nickels. The total value of the jar is $7.85. How many of each type?
Let d be dimes and q be quarters. Set up two equations from our givens:
[LIST=1]
[*]d + q = 41
[*]0.1d + 0.25q = 7.85
[/LIST]
[U]Rearrange (1) by subtracting q from each side:[/U]
(3) d = 41 - q
[U]Now, substitute (3) into (2)[/U]
0.1(41 - q) + 0.25q = 7.85
4.1 - 0.1q + 0.25q = 7.85
[U]Combine q terms[/U]
0.15q + 4.1 = 7.85
[U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.15q%2B4.1%3D7.85&pl=Solve']equation calculator[/URL], we get:[/U]
[B]q = 25[/B]
[U]Substitute q = 25 into (3)[/U]
d = 41 - 25
[B]d = 16[/B]

Kevin and Randy Muise have a jar containing 52 coins, all of which are either quarters or nickels.

Kevin and Randy Muise have a jar containing 52 coins, all of which are either quarters or nickels. The total value of the coins in the jar is $6.20. How many of each type of coin do they have?
Let q be the number of quarters, and n be the number of nickels. We have:
[LIST=1]
[*]n + q = 52
[*]0.05n + 0.25q = 6.20
[/LIST]
We can solve this system of equations three ways:
[LIST]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Substitution']Substitution Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Elimination']Elimination Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Cramers+Method']Cramers Rule[/URL]
[/LIST]
No matter what method we choose, we get the same answer:
[LIST]
[*][B]n = 34[/B]
[*][B]q = 18[/B]
[/LIST]

kim and jason just had business cards made. kim’s printing company charged a one time setup fee of $

kim and jason just had business cards made. kim’s printing company charged a one time setup fee of $8 and then $20 per box of cards. jason,meanwhile ordered his online. they cost $8 per box. there was no setup fee, but he had to pay $20 to have his order shipped to his house. by coincidence, kim and jason ended up spending the same amount on their business cards. how many boxes did each buy? how much did each spend?
Set up Kim's cost function C(b) where b is the number of boxes:
C(b) = Cost per box * number of cards + Setup Fee + Shipping Fee
C(b) = 20c + 8 + 0
Set up Jason's cost function C(b) where b is the number of boxes:
C(b) = Cost per box * number of cards + Setup Fee + Shipping Fee
C(b) = 8c + 0 + 20
Since Kim and Jason spent the same amount, set both cost equations equal to each other:
20c + 8 = 8c + 20
[URL='https://www.mathcelebrity.com/1unk.php?num=20c%2B8%3D8c%2B20&pl=Solve']Type this equation into our search engine[/URL] to solve for c, and we get:
c = 1
How much did they spend? We pick either Kim's or Jason's cost equation since they spent the same, and plug in c = 1:
Kim:
C(1) = 20(1) + 8
C(1) = 20 + 8
C(1) = [B]28
[/B]
Jason:
C(1) = 8(1) + 20
C(1) = 8 + 20
C(1) = [B]28[/B]

Lisa has 32 nickels this is one third of her coins how many coins does she have

Lisa has 32 nickels this is one third of her coins how many coins does she have
Let x be the total amount of coins. we have:
32 = x/3
Cross multiply, we get:
[B]x = 96[/B]

Liz harold has a jar in her office that contains 47 coins. Some are pennies and the rest are dimes.

Liz harold has a jar in her office that contains 47 coins. Some are pennies and the rest are dimes. If the total value of the coins is 2.18, how many of each denomination does she have?
[U]Set up two equations where p is the number of pennies and d is the number of dimes:[/U]
(1) d + p = 47
(2) 0.1d + 0.01p = 2.18
[U]Rearrange (1) into (3) by solving for d[/U]
(3) d = 47 - p
[U]Substitute (3) into (2)[/U]
0.1(47 - p) + 0.01p = 2.18
4.7 - 0.1p + 0.01p = 2.18
[U]Group p terms[/U]
4.7 - 0.09p = 2.18
[U]Add 0.09p to both sides[/U]
0.09p + 2.18 = 4.7
[U]Subtract 2.18 from both sides[/U]
0.09p = 2.52
[U]Divide each side by 0.09[/U]
[B]p = 28[/B]
[U]Now substitute that back into (3)[/U]
d =47 - 28
[B]d = 19[/B]

Lucas has nickels,dimes,and quarters in the ratio 1:3:2. If 10 of Lucas coins are quarters, how many

Lucas has nickels,dimes,and quarters in the ratio 1:3:2. If 10 of Lucas coins are quarters, how many nickels and dimes does Lucas have?
1 + 3 + 2 = 6.
Quarters account for 2/6 which is 1/3 of the total coin count. Let x be the total number of coins. We have:
1/3x = 10
Multiply each side by 3
x = 30
We have the following ratios and totals:
[LIST]
[*]Nickels: 1/6 * 30 = [B]5 nickels[/B]
[*]Dimes: 3/6 * 30 = [B]15 dimes[/B]
[*]Quarters: 2/6 * 30 = [B]10 quarters[/B]
[/LIST]

Marco puts his coins into stacks. Each stack has 10 coins. He makes 17 stacks of quarters. He makes

Marco puts his coins into stacks. Each stack has 10 coins. He makes 17 stacks of quarters. He makes 11 stacks of dimes. He makes 8 stacks of nickels. How much money does Marco have in his stacks of coins?
[U]Value of Quarters:[/U]
Quarter Value = Value per quarter * coins per stack * number of stacks
Quarter Value = 0.25 * 10 * 17
Quarter Value = 42.5
[U]Value of Dimes:[/U]
Dime Value = Value per dime * coins per stack * number of stacks
Dime Value = 0.10 * 10 * 11
Dime Value = 11
[U]Value of Nickels:[/U]
Nickel Value = Value per nickel * coins per stack * number of stacks
Nickel Value = 0.05 * 10 * 8
Nickel Value = 4
[U]Calculate total value of Marco's coin stacks[/U]
Total value of Marco's coin stacks = Quarter Value + Dime Value + Nickel Value
Total value of Marco's coin stacks = 42.5 + 11 + 4
Total value of Marco's coin stacks = [B]57.5[/B]

name five coins that equal 18 cents

name five coins that equal 18 cents
Here are the five coins:
[LIST]
[*][B]1 dime = [/B]10 cents
[*][B]1 nickel = [/B]5 cents
[*][B]3 pennies = [/B]3 cents
[*]5 coins = 10 + 5 + 3 = 18 cents
[/LIST]

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Rob has 40 coins, all dimes and quarters, worth $7.60. How many dimes and how many quarters does he

Rob has 40 coins, all dimes and quarters, worth $7.60. How many dimes and how many quarters does he have?
We have two equations where d is the number of dimes and q is the number of quarters:
[LIST=1]
[*]d + q = 40
[*]0.1d + 0.25q = 7.60
[/LIST]
Using our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=d+%2B+q+%3D+40&term2=0.1d+%2B+0.25q+%3D+7.60&pl=Cramers+Method']simultaneous equation calculator[/URL], we get:
[B]d = 16
q = 24[/B]

Sam has $2.25 in quarters and dimes, and the total number of coins is 12. How many quarters and how

Sam has $2.25 in quarters and dimes, and the total number of coins is 12. How many quarters and how many dimes?
Let d be the number of dimes. Let q be the number of quarters. We're given two equations:
[LIST=1]
[*]0.1d + 0.25q = 2.25
[*]d + q = 12
[/LIST]
We have a simultaneous system of equations. We can solve this 3 ways:
[LIST]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Substitution']Substitution Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Elimination']Elimination Method[/URL]
[*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Cramers+Method']Cramer's Rule[/URL]
[/LIST]
No matter which method we choose, we get the same answer:
[LIST]
[*][B]d = 5[/B]
[*][B]q = 7[/B]
[/LIST]

sample space for flipping a coin 3 times

sample space for flipping a coin 3 times
Each flip gives us 2 possible outcomes, heads or tails. So we have:
2 * 2 * 2 = 8 possible outcomes
[LIST=1]
[*]HHH
[*]HHT
[*]HTH
[*]HTT
[*]THH
[*]THT
[*]TTH
[*]TTT
[/LIST]

Suppose Briley has 10 coins in quarters and dimes and has a total of 1.45. How many of each coin doe

Suppose Briley has 10 coins in quarters and dimes and has a total of 1.45. How many of each coin does she have?
Set up two equations where d is the number of dimes and q is the number of quarters:
(1) d + q = 10
(2) 0.1d + 0.25q = 1.45
Rearrange (1) into (3) to solve for d
(3) d = 10 - q
Now plug (3) into (2)
0.1(10 - q) + 0.25q = 1.45
Multiply through:
1 - 0.1q + 0.25q = 1.45
Combine q terms
0.15q + 1 = 1.45
Subtract 1 from each side
0.15q = 0.45
Divide each side by 0.15
[B]q = 3[/B]
Plug our q = 3 value into (3)
d = 10 - 3
[B]d = 7[/B]

the sample space for a coin being tossed twice

the sample space for a coin being tossed twice
Since each toss results in 2 outcomes, we have 2^2 = 4 possible events in the sample space:
[LIST=1]
[*]H,H
[*]H,T
[*]T,H
[*]T,T
[/LIST]

there are $4.20 in nickel and quarters. There are 6 more nickels than quarters there. How many coins

there are $4.20 in nickel and quarters. There are 6 more nickels than quarters there. How many coins of each are there
We're given two equations:
[LIST=1]
[*]n = q + 6
[*]0.05n + 0.25q = 4.2
[/LIST]
Substitute equation (1) into equation (2):
0.05(q + 6) + 0.25q = 4.2
Multiply through and simplify:
0.05q + 0.3 + 0.25q
0.3q + 0.3 = 4.2
To solve for q, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.3q%2B0.3%3D4.2&pl=Solve']type this equation into the search engine[/URL] and we get:
q = [B]13
[/B]
To solve for n, we plug in q = 13 into equation (1):
n = 13 + 6
n = [B]19[/B]

three coins are tossed.how many different ways can they fall?

three coins are tossed.how many different ways can they fall?
[URL='https://www.mathcelebrity.com/cointoss.php?hts=+HTHTHH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=heads&tossct=3&montect=3&calc=5&pl=Calculate+Probability']8 outcomes using our coin toss calculator[/URL]

Two coins are flipped 2 times. Calculate the total outcomes of these coins.

Two coins are flipped 2 times. Calculate the total outcomes of these coins.
2 coins * 2 outcomes per coin = 4 possible outcomes
[LIST=1]
[*][B]H,H[/B]
[*][B]H,T[/B]
[*][B]T,H[/B]
[*][B]T,T[/B]
[/LIST]

what is the probabilty of tossing two coins and both landing on heads

what is the probabilty of tossing two coins and both landing on heads
We want P(HH). We type in [URL='https://www.mathcelebrity.com/cointoss.php?hts=HH&hct=+2&tct=+1&fct=+5>=no+more+than&nmnl=+2&htpick=tails&calc=1&montect=500&pl=Calculate+Probability']HH into our search engine[/URL] and we get:
P(HH) = [B]0.25 or 1/4[/B]

You have 4 dimes, 1 quarter and 6 pennies. How many cents do you have? Write it as a decimal

You have 4 dimes, 1 quarter and 6 pennies. How many cents do you have? Write it as a decimal
We type in [URL='https://www.mathcelebrity.com/coinvalue.php?p=6&n=&d=4&q=1&h=&dol=&pl=Calculate+Coin+Value']4 dimes, 1 quarter, 6 pennies into our search engine[/URL] and we get:
[B]0.71 as a decimal for cents[/B]

You want to put 520 quarters in coin wrappers. You need one wrapper for every $10 in quarters. Write

You want to put 520 quarters in coin wrappers. You need one wrapper for every $10 in quarters. Write an equation you can use to find how many wrappers w you need
First, calculate the number of quarters in $10:
Quarters in $10 = Value of Quarters / Cost per quarter
Quarters in $10 = 10/0.25
Quarters in $10 = 40
Now find out how many wrappers we need with each wrapper holding 40 quarters:
Number of wrappers = Total quarters / Quarters per wrapper
Number of wrappers =520/40
Number of wrappers = [B]13[/B]