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3 years ago

What is the answer!!! 35.00 for a jacket. 10% discount 5% sales tax whats the final cost? PLEASE HELP ME

1 answer:

6 0

Firstly, you should find the discount that is taken off of the jacket, and then also find the sales tax.
As the discount is 10%, you simply have to divide 35 by 10. 35/10= 3.50.
Therefore, you need to subtract the 3.50 from the 35, which gives you 31.50. You've then got the 5% sales tax to add on, which is 3.50/2= 1.75, you've then got to add 1.75 to 31.50, which gives you 33.25.
However, if the sales tax is on the discounted price, then it will be 31.50/10= 3.15/2= 1.58, and you've then got to add this is to discount price, which is 31.50+1.58= 33.08

Hope this helps :)

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Step-by-step explanation:

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3 years ago

Simplify the expression below: 4+2×[(3+1)×5] (show work please need it for review)

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Answer:

Step-by-step explanation:

Parenthesis

Exponents

Multiply

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Add

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First we need to do the work inside the parenthesis that are in the boxed parenthesis

3 + 1 = 4

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Now we multiply

20 * 2 = 40

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4 years ago

X=8<br> what is the radical equation

kap26 [50]

Answer:

x=64

Step-by-step explanation:

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3 years ago

Which of the following is the solution to the

Studentka2010 [4]

Answer:

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2 years ago

Jeanne has many nickels, dimes, and quarters in her wallet. She chooses 3 coins at random. What is the probability that all thre

Whitepunk [10]

Answer:

Step-by-step explanation:

There isn't enough said about the distribution of coins in her wallet, but we'll just assume that the number is so large that any coin is equally likely to be drawn.

Stated another way, there are 27 possible outcomes of the three draws (3 x 3 x 3) and we'll assume each is equally likely.

PROBLEM 1:

This is a conditional probability question. We only have to consider the cases where she could have drawn 2 quarters and another coin. The possible draws are:

DQQ, NQQ, QDQ, QNQ, QQD, QQN or QQQ*.

That's 7 possible draws (with equal probability) and only 1* of them is a draw with 3 quarters.

Answer:

P(three quarters given two are quarters) = 1/7

PROBLEM 2:

Again, this is conditional probability. To help count the ways, let's instead count the ways to *not* draw any dimes. That means you have 2 choices for the first coin, 2 choices for the second coin and 2 choices for the third coin.

So 8 out of the 27 draws would *not* contain a dime. By subtracting, we can see that 19 of the draws *would* contain at least one dime.

Now think of the ways to create a draw consisting of one of each coin. We have the 3 different coins and they can be drawn in any order. That would be 3! or 6 ways.

If that isn't clear, let's list them all out:

DDD, DDN, DDQ, DND, DNN, DNQ*, DQD, DQN*, DQQ, NDD, NDN, NDQ*, NND, NQD*, QDD, QDN*, QDQ, QND*, QQD

There are 19 possible outcomes with at least one dime and exactly 6 of them have one of each type.

P(all different given at least one is a dime) = 6/19

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3 years ago