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ankoles [38]
2 years ago
7

A piggy bank contains some dimes and nickels. There are 8 more dimes than nickels in the bank. There is a total of $1.40. How ma

ny of each type of coin are in the bank?
Mathematics
1 answer:
Softa [21]2 years ago
5 0

Answer:

4 nickels

12 dimes

Step-by-step explanation:

Dimes are worth .1 each while nickels are .05 each.

We have 8 more dimes than nickels. Let d represent number of dimes and n for number of nickels. This means we have d=8+n.

If all our nickels and dimes together are worth 1.4 then we have another equation .1d+.05n=1.4

Lets put our equations together:

d=8+n

.1d+.05n=1.4

‐-----------------Plug first equation into second.

.1(8+n)+.05n=1.4

Distribute

.8+.1n+.05n=1.4

Combine like terms

.8+.15n=1.4

Subtract .8 on both sides

.15n=.6

Divide both sides by .15

n=.6/.15=4

Remember there are 8 more dimes so d=8+4=12.

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lesya692 [45]
Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes P_B(A).

The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
P(A|B)= \frac{P(A \cap B)}{P(B)}

In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.

The probability of selecting one coin is \frac{1}{3}

Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.

P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.

Thus P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}

</span><span>P(B) means that the first envelope contains a quarter AND the second envelope contains a quarter

</span><span>Thus P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}

Therefore, P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}


Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}

</span><span>P(D)= \frac{1}{3}</span><span>

Therefore, P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}</span>
3 0
2 years ago
A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15.
mariarad [96]

Answer:

a. the probability that any one of the computers will require repair on a given day is constant

Step-by-step explanation:

The following properties must be true in order for a distribution to be binomial:

- A fixed number of trials (125 computers)

- Each trial is independent of the others (one computer requiring repair does not interfere with the likelihood of another requiring repair).

There are only two outcomes (requires repair or do not require repair)

The probability of each outcome remains constant from trial to trial (All computers have the same likelihood of requiring repair, 0.15).

Therefore, the alternative that better fits those properties is alternative a. the probability that any one of the computers will require repair on a given day is constant

8 0
2 years ago
Below is a scale drawing of the town swimming pool in the drawing the longest side of the pool has length of 10 inches Paco crea
quester [9]

Answer:

2:1 that is 2 inches in real life is equal to 1 inch in the drawing.

Step-by-step explanation:

We are given the following in the question:

Longest side of the pool = 10 inches

Longest side of the pool in drawing = 5 inches

Scale:

We find the ratio of the real length of pool and the measurement of the length of the pool in the drawing.

\text{Scale} = \dfrac{10\text{ inches}}{5\text{ inches}} = 2:1

Thus, we can say Paco's new drawing follows a scale of 2:1 that is 2 inches in real life is equal to 1 inch in the drawing.

6 0
2 years ago
Read 2 more answers
Admission for a water park is 17.50 per day.a season pass cost 125. How many days must you go to the water park to make buying t
solmaris [256]

You must go to the water park for 8 days to make buying the season pass worth it.

<em><u>Explanation</u></em>

Suppose, the number of days you must go to the water park is  x

Admission for the park is 17.50 per day. So, <u>the total cost for x days</u> will be:  17.50x

Given that, the season pass cost 125. So, the equation will be.....

17.50x=125\\ \\ x=\frac{125}{17.50}=7.142...\\ \\ x \approx 8

<em>(As the number of days can't be in decimal)</em>

So, you must go to the water park for 8 days to make buying the season pass worth it.

7 0
2 years ago
The data plan for Shawn's phone has 32,000 megabytes of storage for
olga nikolaevna [1]

Answer:

Yes she 1280 megabytes of additional 4% data and requires 1000 megabytes to upload 200 photos

Step-by-step explanation:

Given:

Data Plan for Shawn's Phone = 32000 megabytes.

She want to increase the data by 4% = \frac{4}{32000}\times 100= 1280 megabytes

Each Photo uses 5 megabytes so 200 additional Photo will require =5\times200= 1000\ megabytes

She has 1280 megabytes of additional data and requires 1000 megabytes to upload 200 photos, so she have enough memory to do so.

4 0
2 years ago
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