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Amanda [17]
4 years ago
15

twenty of the students in Hannah's class, or 80% of the class, voted to have pizza for lunch every Wednesday. How many students

are in Hannah's class?
Mathematics
1 answer:
Igoryamba4 years ago
6 0

Answer:

There are 25 students in Hannah's class.

Step-by-step explanation:

If 20 students in Hannah's class is 80% of the total amount of students, we can use a variable to figure out the total amount of students.

Let's say x is the total amount of students. We can use this equation to solve the problem:

20 = .8x

This is saying 20 equals 80% (.8 when you move the decimal 2 times) of x, or the total amount of students. Let's solve.

20 = .8x

Divide by .8

25 = x

There are 25 students in Hannah's class.

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1 Pound brown sugar is equivalent to _____ cups
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3 1/2 or said as 3.5 cups is equal to 1 pound

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Can someone please answer this question I need it today please show work and please answer it correctly i
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Answer:

24) $495

25) 14%

26) 25/X = 83/100

27) 0.7p

28) x + .085x and 1.085x

29) $221.90

30) $24.10

31) $6.13

32) 40%

Step-by-step explanation:

24) 600 - (600 × 0.25) = 450

450 × 1.10 = 495

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29) 100% - 15% = 85%

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4 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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