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guajiro [1.7K]
3 years ago
9

A set of 10 coins may contain any combination of pennies, nickels, dimes, quarters, or half-dollars. In how many different ways

can the set of 10 coins have a total value of 59 cents?
Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

that is a very complicated question with alot of variable to figure out

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Solve the following formula for m<br><br> C = amt<br><br> m =<br><br> Answer as expression
d1i1m1o1n [39]
M=c/at

Divide both sides by a and t to isolate m. Then the solution is just m=c/at
4 0
3 years ago
PLEASE HELP DUE IN 20 MINUTES FIND X AND SHOW YOUR WORK PLEASE
Vlad1618 [11]

Answer:

x = 7°

Step-by-step explanation:

3x° - 53° = x° + 67°

2x° = 14°

x° = 7°

3 0
2 years ago
One of ten different prizes was randomly put into each box of a cereal. If a family decided to buy this cereal until it obtained
Reptile [31]

Answer:

29.29

Step-by-step explanation:

The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.

The expected value of a negative binomial is the reciprocal of its success probability <em>p.</em>

<em>E(X)=</em>1/<em>p</em>

At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success

<em>E(X)=</em>1/<em>p</em>

<em>E(X1)</em>=1

After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (<em>p</em>=9/10)

E(X2)=1/9÷10

E(X2)= 10/9

After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

E(X3)=1/8÷10

E(X3)=10/8

E(X3)=5/4

After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

E(X4)=1/7÷10

E(X4)=10/7

Fifth different prize

E(X5)=1/6÷10

E(X5)=10/6

E(X5)=5/3

Sixth different prize

E(X6)=1/5÷10

E(X6)=10/5

E(X6)=2

Seventh different prize

E(X7)=1/4÷10

E(X7)=10/4

E(X7)=5/2

Eighth different prize

E(X8)=1/3÷10

E(X8)=10/3

Ninth different prize

E(X9)=1/2÷10

E(X9)=10/2

E(X9)=5

After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize

E(X10)=1/1÷10

E(X10)=10

Add up all corresponding values;

<em>E(X)=</em>E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)

<em>E(X)=</em>1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10

<em>E(X)</em>= 29.29

Note: Those values in <em>E(X), </em>should be written the way it will in standard algebra ie they should be small.

4 0
3 years ago
Juan pulled a red card from the deck of regular playing cards. This probability is 2652 or 12 . He puts the card back into the d
kenny6666 [7]

Answer:

no

Step-by-step explanation:

if he randomly pulls it out it's still the same number of red cards and also the total number of cards as always

can't change

6 0
3 years ago
Read 2 more answers
Find the value of x. Round the length to the nearest tenth.
attashe74 [19]

I donno is there anything else

8 0
3 years ago
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