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laiz [17]
1 year ago
7

The price of a ring was increased by 30% to £325. What was the price before the increase?

Mathematics
2 answers:
Oxana [17]1 year ago
7 0

Answer:

The cost before the increase is £250.

Step-by-step explanation:

Given that after 30% increases, the cost of the ring is £325 so we can assume that 130% is £325. Now, we have to form an expression in term of x where x represents the original cost :

\frac{130}{100}  \times x = 325

Next, we have to solve it :

x = 325 \div  \frac{130}{100}

x = 325 \times  \frac{100}{130}

x = 250

Radda [10]1 year ago
3 0

Answer:

227.5

Step-by-step explanation:

Find the number that is increased by 30% to get to 325.  

100-30 = 70.

In other words, what number is 70% of 325.

325 x .70 = 227.5

The original price of the ring was 227.5.

Check your work, 227.5 = 70% of X

227.5  / .70  = 325

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2 years ago
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Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
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Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

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The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145

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2 years ago
is x + 10 a factor of the function f(x) = x3 − 75x + 250? Explain. (2 points) Yes. When the function f(x) = x3 − 75x + 250 is di
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Asked and answered elsewhere.
brainly.com/question/10474342

Your appropriate choice is ...
  Yes. When the function f(x) = x3 − 75x + 250 is divided by x + 10, the remainder is zero. Therefore, x + 10 is a factor of f(x) = x3 − 75x + 250.
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