Can someone help me with this problem...
#20
Can someone help me with this problem... #20 - 1

Answers

Answer 1
Answer: 3/12, 1/3, 2/5, 4/9, 5/8, 2/3, 3/4, 5/6

Related Questions

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In the given figure , if PQ ll RS , angle TRS = 105 and angle RTP = 30 , then the value of x is
If 15% is $375 how much is 5%
What's 5 5/6 as a improper fraction
Y=5x Is it proportional or not ?

The table below shows the number of marbles of different colors in a bag:Marble Experiment



Color of
Marbles

Number of
Marbles



Red

5


White

8


Black

2



Deja draws a marble from the bag randomly without looking. She then draws another marble from the bag without replacing the first one. Which expression shows the probability of drawing red marbles in both the trials?


5 over 15 multiplied by 4 over 14
5 over 15 multiplied by 4 over 15
5 over 15 added to 4 over 14
5 over 15 added to 4 over 15

Vincent is giving cakes to some children at a carnival. He has 2 strawberry cakes, 6 pineapple cakes, and 7 chocolate cakes. If Vincent selects a cake randomly without looking, what is the probability that he will give a pineapple cake to the first child and then a strawberry cake to the second child?

6 over 15 multiplied by 2 over 14 is equal to 12 over 210
6 over 15 plus 2 over 14 is equal to 114 over 210
6 over 15 multiplied by 2 over 15 is equal to 12 over 225
6 over 15 plus 2 over 15 is equal to 8 over 15

Answers

There are a total of 5+8+2=15 marbles.  There's a 5/15 chance of drawing a red marble in the first trial.  Then, there are 4 red marbles and 14 total marbles left, so afterwards, there's a probability of 4/14 that the second marble is red.  The probability overall is found by multiplying these together, so the first answer is the best.

There are 2+6+7=15 cakes.  There's a 6/15 chance that the first cake selected will be a pineapple cake, and then a 2/14 chance that the second cake selected will be a strawberry cake, by the same logic as above.  This is equal to 12/210.  The first answer is the best option.

An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that Casts a Shadow of 15 metres under similar conditions. ​

Answers

Answer: The height of the tree will be 21 meters

How do you find the height of a 3d object such as a triangle or a cylinder

Answers

In the case of a cylinder, one or several of the following data must be provided: the total area/lateral area and the circumference/area of the base or the area of the base and the volume. Ex: provided the circumference of the base = 10π inches, and the lateral area = 50π inches², the equation can be written as 50π ÷ 10π = 5 inches.
Well, for ANY prism (included cylinder), divide volume by area of base.
h = V / B

For ANY pyramid (included cone), divide volume by area of base, then multiply it by 3.
h = 3V / B

How do you simplify a fraction

Answers

To simplify a fraction, you must find a number both the numerator and denominator can be divided into, and divide. Say I have 4/6. Both numbers are divisible by 2, so I simplify the fraction to 2/3. As we can no longer divide, the fraction is in simplest form. Hope this helps!
say that the fraction is 6\28. what number can be divided into 6 and 28? 2 because 28 divided by 2 equals 14 and 6 divided by 2 equals 3. So then you have 3\14.

The sales tax rate is 8%. If Jim bought a new Buick and paid a sales tax of $1,930, what was the cost of the Buick before the tax?

Answers

Let the price of the car = x.

8% of the price of the car is $1930.

We now translate the sentence into an equation using x as the unknown amount. Then we solve the equation to find x, the price of the car before tax.

8% * x = 1930

0.08x = 1930

x = 1930/0.08

x = 24,125

Answer: The price of the car before tax is $24,125.

David says you can have an unlimited number of equivalent fractions to any given fraction. Is he right?

Answers

Yes, David is correct.
We can test our hypothesis by setting up an experiment.
1   2    4   6    7
2   4    8   16  14
AND SO ON!
yes, multiply the numerator and denominator by the same number again and again to keep getting different equivalent fractions.