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3 years ago

I need help so can u help me ok here is the problem write a story problem with 6 divided by 3/4

Mathematics

2 answers:

Lostsunrise [7]3 years ago

5 0

0.5 is the answer for the problem

never [62]3 years ago

3 0

Here is an optional story problem:

You have a ribbon that is 6 inches long.
You need to cut sections that are each 3/4 inches long. How many sections of ribbons will you have?

The answer to this is 8 sections of ribbon.

Hope this helped!

Comment if you have any questions!!!

You might be interested in

Find the probability of rolling an odd sum or a sum less than 7 when a pair of dice is rolled

Nutka1998 [239]

Answer:

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

$P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818$

Step-by-step explanation:

For this case when a pair of dice is rolled we have the following sample pace for the outcomes:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7

For the pairs that satisfy that the sum is less than 7 we have:

(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)

So we have a total of 15 pairs where the sum is less than 7

And for an odd sum we have the following pairs

(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)

A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:

$P = \frac{15+18-6}{36}= \frac{27}{33}= 0.818$

7 0

3 years ago

Help asap plssss, brainliest!!

liubo4ka [24]

Answer:

Step-by-step explanation:

85+ 80 = 185

i dont think this is right tho

6 0

3 years ago

Read 2 more answers

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>

5 0

2 years ago

How do you solve common core math when it asks you to write the first step question and answer?

slega [8]

Represent addition and subtraction with objects, fingers, mental images, drawings1, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.

7 0

3 years ago

HURRY PLEASE!<br> (g+2)^5

KengaRu [80]

Step-by-step explanation:

We need to use the binomial theorem/Pascal's triangle here.

(a+b)^5 = (5 choose 0)a^5 + (5 choose 1)a^4*b + (5 choose 2)a^3*b^2 + (5 choose 3)a^2*b^3 + (5 choose 4)a*b^4 + (5 choose 5)b^5.

5 choose 0 = 1

5 choose 1 = 5

5 choose 2 = 10

5 choose 3 = 10

5 choose 4 = 5

5 choose 5 = 1

And 1, 5, 10, 10, 5, 1, is the (5+1) = 6th row of pascal's triangle.

Therefore we get

g^5 + 5g^4*2 + 10g^3*2^2 + 10g^2*2^3 + 5g*2^4 + 2^5

which is

g^4 + 10g^4 + 40g^3 + 80g^2 + 80g + 32

Or, you could do the slow way, by just doing (g+2)(g+2)(g+2)(g+2)(g+2)

3 0

3 years ago