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Pepsi [2]
3 years ago
11

Pls help 6th grade math ill be adding brainlist

Mathematics
2 answers:
iren2701 [21]3 years ago
4 0

Answer:

300 is the answer as only 27% is given we calculated of 100%.

bagirrra123 [75]3 years ago
4 0

1. 22 students is 40% of 55

2. 24 red marbles is 40% of  60 marbles

3. 15% of $9 is $1.35

4. 12 is 6% of 200

5. Bethany sent 9 text messages to her friend yesterday.

6. There where 17.01 people in the survey.

7. The sales tax rate is 6.5%.

8. There were 10(0.1) from 18-29 that voted in the election.

9. There were 35(0.35) from 50-64 that voted in the election.

10. There were 25(0.25) of over 64 year old's that voted in the election.

11. Sahli  is incorrect because there is 30 people within ages 30-49. I think he came up with that answer so quickly because 30 + 49 = 1470 and divided by 14 = 105 with this you add 30 and get 135.

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3 more than a number then divided the result by 8
ziro4ka [17]

We are given statement : 3 more than a number then divided the result by 8.

We need to write an algebraic expression for it.

Let us assume unknown number be n.

3 more than n = (n+3).

Now, we need to divide that result (n+3) by 8.

So, we would get (n+3) divided by 8 = \frac{n+3}{8}.

<h3>Therefore, final expression is \frac{n+3}{8}.</h3>
5 0
3 years ago
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A basin has the shape of a prism. What is the capacity of this basin? *
dolphi86 [110]
I think it’s c.....
3 0
3 years ago
20. The surface areas of two similar solids are 216 m² and 1014 m². The volume of the larger one is 2197 m³. What is the volume
Andrei [34K]

Answer:

  216 m³

Step-by-step explanation:

The ratio of linear dimensions is the square root of the ratio of area dimensions.

  s = √(216/1014) = √(36/169) = 6/13

Then the ratio of volume dimensions is the cube of that. The smaller volume is ...

  v = (6/13)³·2197 m³ = 216/2197·2197 m³ = 216 m³

The volume of the smaller solid is 216 m³.

8 0
3 years ago
Help could it be any two integers?
Novosadov [1.4K]

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52) -1 and -3

53) -9 and -10

5 0
3 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
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