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Allushta [10]
3 years ago
6

Please help ASAP! i am having issues!

Mathematics
2 answers:
adell [148]3 years ago
3 0

i believe the answer is D.) 1 over 8

Explanation: 4 is one of the 8 numbers so it would be a 1 over 8 chance.

inysia [295]3 years ago
3 0
I believe that it is d 1/8
You might be interested in
The image shows three tennis balls enclosed in a cylindrical can. Choose all that are correct. The volume of a single tennis bal
Fynjy0 [20]

Answer:

The volume of a single tennis ball is 14.14\ in^{3}

The volume of three tennis balls is 42.42\ in^{3}

Step-by-step explanation:

Step 1

Find the volume of a single tennis ball

we know that

The volume of a sphere is equal to

V=\frac{4}{3}\pi r^{3}

where r is the radius of the sphere

In this problem

r=3/2=1.5\ in

substitute

V=\frac{4}{3}\pi (1.5)^{3}

V=14.14\ in^{3}

Step 2

Find the volume of the can

we know that

the volume of the cylinder is equal to

V=\pi r^{2} h

where r is the radius of the cylinder

h is the height of the cylinder

in this problem we have

r=3/2=1.5\ in

h=8.4\ in

substitute

V=\pi (1.5)^{2}(8.4)

V=59.38\ in^{3}

Statement

<u>case A)</u> The volume of a single tennis ball is 14.14\ in^{3}

The statement is true

See the procedure in Step 1

<u>case B)</u> The volume of the can is 56.55\ in^{3}

The statement is false

The volume of the can is 59.38\ in^{3} --> see the procedure Step 2

<u>case C)</u> The empty space inside the can is 14.13\ in^{3}

The statement is false

To find the empty space subtract the volume of three tennis ball from the volume of the can

59.38\ in^{3}-3*14.14\ in^{3}=16.96\ in^{3}

<u>case D)</u> The volume of three tennis balls is 42.42\ in^{3}

The statement is true

To find the volume of three tennis balls multiply the volume of a single tennis ball by three

14.14*3=42.42\ in^{3}

8 0
3 years ago
3/6 + 1/2 what is the answer.​
Simora [160]

Answer:

reduce 3/6 to lowest terms which would be, 1/2 because 3 goes into 3 once and 3 goes into 6 twice.  

Making your equation simplified to

1/2 + 1/2

which is equal to 1

1/2 + 1/2 = 1

Step-by-step explanation:

Hope this helps! Have a fantastic day and please mark brainliest

4 0
3 years ago
Read 2 more answers
A mine elevator was at a depth of -120 feet. After 15 seconds, the mine elevator was at a depth of -30 feet. Which expression sh
KonstantinChe [14]
-120 + -30 = 90. 90/15 = 6
Answer ; 6mph
8 0
3 years ago
Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair
Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

 

 

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102.. 

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

 

 

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

 

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

 

3*5: numbers with common factors >1, with 3*5 must be 

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

 

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

 

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....}  containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

 

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

 

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0
3 years ago
Round to the nearest cent $6.51
son4ous [18]
Answer: $6.50
If the number is below 5 (so 1-4) you round down. If its above 5 (5-9), you round up.
E.G. 78.89 
Round this to the nearest 10th, you'll get 78.90

4 0
3 years ago
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