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nikdorinn [45]
2 years ago
8

Eighty, decreased by three times a number, is the same as five times the number, increased by eight

Mathematics
2 answers:
OlgaM077 [116]2 years ago
4 0
Idk if your looking for a solution or an equation but the equation is 80-3x = 5x
pickupchik [31]2 years ago
3 0

This is your equation: :)

80-3x=5x+8

Answer:

x=9

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The value of which of these expressions is closest to e?
pishuonlain [190]
The correct answer for the exercise shown above, is the first option (Option A), which is:

 A. <span>(1+1/18)^18

 The explanation is shown below:

 1- You have that the number e is an irrational number, therefore, its value is aproximated:

 e=2.7182

 2- You have the following expressions:

 </span>(1+1/18)^18=2.6464 
 <span>(1+1/17)^17=2.6424
 </span><span>(1+1/16)^16=2.6379
 </span><span>(1+1/15)^15=2.6328

 Therefore the value of the expression that is closest to e is the Option A.</span>
5 0
2 years ago
What key would you use if 10 students chose cartoons
NISA [10]

I am not 100% sure as to what you are asking, but I will try my best. Each key is equal to 3 students, so if 10 chose cartoons, a better key would be 2, 5, and 1. 1 might make it a bit cluttered, but if 10 students chose cartoons and you are using the 3 key, you can draw 1/3 of a box.

4 0
2 years ago
Read 2 more answers
If G is midpoint of FH, find FG
pochemuha
The answer is 8 ihinghkki
8 0
2 years ago
Giuliana has 22 quarters and dollar coins worth a total of $10.75. Which equation can be used to find d, the number of dollar co
vovangra [49]

Answer:

Option (3). 0.25(22 - d) + d = 10.75

Step-by-step explanation:

Total number of coins Giuliana has = 22

Let the number dollar coins Giuliana has = d

Number of quarters with Giuliana = (22 - d)

Value of dollar coins and quarters = $10.75

So the equation will be,

Number of dollars coins × $1 + Number of quarters × $0.25 = $10.75

d + 0.25 × (22 - d)  = 10.75

Therefore, to determine the number of dollars with Giuliana equation will be used,

d + 0.25(22 - d) = 10.75

Option (3) will be the answer.

4 0
2 years ago
Read 2 more answers
Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang
kkurt [141]

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:

P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164

c) We can repeat the method for the previous probabilty.

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

5 0
2 years ago
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