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OverLord2011 [107]
3 years ago
7

the ratio of boys to girl in Mr. Claytons class is 2:3. If there are 30 total students how many more girls are in the class than

boys
Mathematics
2 answers:
Dafna11 [192]3 years ago
6 0
Well, if there are two boys for every three girls then we multiply 30 by 3.

30×5=90.

Next, we divide 90 by 5.

90÷5=18.

Then we subtract 18 from 30.

30-18=12.

That means we have 12 boys and 18 girls.

Finally, we subtract 18 from 12.

18-12=6.

There are 6 more boys than girls.

Have a nice day! :)


pashok25 [27]3 years ago
3 0
Number of girls :
30*3/5=18

Number of boys :
30*2/5=12

18-12=6

6 more girls than boys.
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Answer:

1. 9/10  

2. 14/15

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Step-by-step explanation:

1. Change the 5, the denominator to 10.  8/10

then add to 1/10  8/10 + 1/10= 9/10

2. Change both denimnators to 15. You will then get 9/15 and 5/15.

Add them together and get 14/15

3. Change both denominators to 6 and you will get 3/6 and 4/6.

Add them together to get 7/6.

If you want to make it into a mixed fraction it would be 1 1/6 (as in one and one-sixth)

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Answer:

EF = 6.6

Step-by-step explanation:

Since ABCD is similar to EFGH, then EH is similar to AD. So, we can solve by first dividing 12 by 2 (EH by AD). The quotient of this is 6. This tells us that quadrilateral EFGH is 6 times larger than quadrilateral ABCD, since they are similar. So, with this and the measurement of AB (which is similar to EF), we can now solve for EF. We simply multiply 1.1 (the measurement of AB) by 6 (how many times larger EFGH is compared to ABCD). The product of this is 6.6, our final answer.

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X represents the number that would be added to the silver cars.


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3 years ago
A cold drink is poured out at 52°F. After 2 minutes of sitting in a 72°F room, its temperature has risen to 55°F. Find an equati
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Answer:

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

Step-by-step explanation:

For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

\frac{dT}{dt}=k*(T_r-T)

If we rearrange and integrate

\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}

We know that at time 0, the temperature of the drink was 52°F. Then we have:

T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20

We also know that at t=2, T=55°F

T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08

The model for the temperature of the drink can be written as

T=72-20e^{-0.08t}

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