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s2008m [1.1K]
3 years ago
5

• A photo store charges $1 for a print of a 5-inch * 7-inch photo, and $2 for a print

Mathematics
2 answers:
Lerok [7]3 years ago
6 0

Answer:

im not sure try to see examples of this question, and maybe it can help you find the answer. sorry i cant help

Step-by-step explanation:

NARA [144]3 years ago
4 0

Answer:

Step-by-step explanation:

I'm pretty sure the answer is The cost of any number of 8x10 photos is twice the cost of the same number of 5x7 photos. Just by looking at the chart. Hope this helps

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There is a markup of 40% on the original price of a fan. If the fan was sold for $42, what was its original price
Akimi4 [234]

Answer:   21$ that is the answer

Step-by-step explanation:

4 0
3 years ago
A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2
Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.                  

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?    

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is \frac{1}{20000}

Probability of getting second prize is \frac{2}{20000}

Probability of getting third prize is \frac{20}{20000}

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is \frac{19977}{20000}

The expected value is

E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25

E=\frac{1000+600+200-4994.25}{20000}

E=\frac{-3194.25}{20000}

E=-0.159

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0
3 years ago
Show full workout <br>Change the following bases to decimal<br>a. 101011<br>b. 2373​
Lubov Fominskaja [6]
B I think if I’m wrong I sooooo sorry
4 0
3 years ago
Please help me branniest
Julli [10]

Answer:

just add the powers to get 6 to the 10 power


3 0
3 years ago
An electronic store prices 3 dozen video games at a total of 1467 if each video game costs the same amount what m is the total c
PilotLPTM [1.2K]

Answer:4 dozens of video games would cost $1956

Step-by-step explanation:

An electronic store prices 3 dozen video games at a total of 1467. A dozen of video games is 12. Therefore, the total number of video games whose cost is 1467 is 12 × 3 = 36

if each video game costs the same amount, then the cost of one video game would be

1467/36 = $40.75

The total number of video games in 4 dozens of video game would be

4 × 12 = 48

Therefore, the total cost of 48 video games would be

48 × 40.75 = $1956

4 0
3 years ago
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