When Andy goes to the shop ,he finds that: The regular price of a single candy bar is 79 cents.
Packs of 5 candy bar cost $3.
There is a special offer of “buy two get one free”.

Question 1.find the cost of buying 20 single candy bars at the regular price.
Question 2.how much will Andy pay if he buys 20 candy bars in packs of 5?
Question 3.find the cost of buying 20 candy bars at the special price.

1. $15.80

2. $12

3. $11.06

Step-by-step explanation:

Step-by-step explanation:

1. Buying 20 single candy bars at regular price

The regular price of one is 79 cents.

(# of bars) X (price per bar) = regular price

20 X 79 cents = 1580 cents

1580 cents = $15.80

2. Buying 20 bars in packs of 5

Each pack is $3

(# of packs) X (price per pack) = pack price

(20/5) X ($3) = pack price

4 X $3 = $12

3. Buying 20 bars at special price

For every 2 bought, 1 is free.

(Number of bars paid for) X (price per bar) = special price

(2(20/3)) X (0.79) = special price

14 X 0.79 = $11.06

Because for every 2 bars that are paid for, 1 is free, he pays for 14 bars and he gets the 21st free.

Andy's 20 regular priced candy bars are $15.80, bars in packs are $12, and bars at the special price is $11.06.

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.