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3 years ago

6

100% & There are 16 girls and 18 boys in a class. The teacher chooses a student's name at What is the probabilit

y that the teacher chooses a girl to answer the question?
A.
34
O
B.
16
o
C.
8
17
OD.
Ol

2 answers:

Mariulka [41]3 years ago

8 0

Answer:a

Step-by-step explanation:

djyliett [7]3 years ago

8 0

Answer:

my answer iA. because 16+18 =34

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El volumen en L, si 6,35 mol tiene una presión de 2,7 atm a 4 °C.

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Answer:

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3 years ago

Need help............

svp [43]

Answer:

i would say C

Step-by-step explanation:

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3 years ago

There is a decrease in power. It can be concluded that there has also been a change in: A)current or voltage.

Serhud [2]

There is a decrease in power. It can be concluded that there has also been a change in:

A) current or voltage.

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3 years ago

The wind force f on a sail varies jointly as the area a of the sail and the square of the wind speed w.

blondinia [14]

Answer:

122.5 lbs

Step-by-step explanation:

The relation between force, wind speed, and area can be written ...

f = ks²a

for some wind speed s and area a.

__

The constant of proportionality, k, can be found from the given data.

64.8 = k(18²)(500) . . . force in pounds, area in square feet, speed in mph

k = 64.8/(324×500) = 0.0004

__

Using the equation with the second set of data, we have ...

f = 0.0004(35²)(250) = 122.5 . . . . pounds

The force on the 250 square foot sail would be 122.5 pounds.

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2 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.

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2 years ago