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Readme [11.4K]
3 years ago
7

There are 80 sixth graders at Wilson Middle School. Only 70% of the sixth graders will attend the morning assembly. How many six

th graders will be at the morning assembly?
Mathematics
2 answers:
Nuetrik [128]3 years ago
6 0

The answer is 56 because

Recall that 70% is , and  is 0.70.


To find the number of sixth graders who will attend the morning assembly, multiply 80 by , or 0.70.




So, 56 sixth grade students will attend the morning assembly.[?][?][?][?][?][?][?][?]


zysi [14]3 years ago
4 0
1 student will attend the assembly in the morning
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lina2011 [118]
By their increase or decrease
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3 years ago
Shamika's grandmother opens an account with a deposit of $5,000 for Shamika as a college savings account. The account pays 5 per
lesantik [10]

Answer:

$7500

Step-by-step explanation:

$5000 was deposited

5% annual interest of the $5000 is $250

so in ten years the interest will be 10 ×$250 =$2500

so the total sum that will be in the account is $5000 + $2500 =$7500

7 0
3 years ago
What two whole numbers is the quotient between?
Katarina [22]

Answer:

Choice D

Step-by-step explanation:

38/4 =9.5

Therefore, this number is between 9 and 10

Note:

38/4 = 9 remainder 2

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3 0
3 years ago
A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
3. List all the coefficients and constants
7nadin3 [17]
The coefficient is 4
The constants are 2 and 3
5 0
3 years ago
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