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Ksenya-84 [330]

3 years ago

8

On Monday, a security guard spent 4.5 hours on patrol and 2 hours on surveillance. On Tuesday, the officer spent 1 hour and 40 m

inutes on surveillance and 3 hours on patrol. Which day has a greater ratio of time on surveillance to time on patrol

1 answer:

Troyanec [42]3 years ago

6 0

The first one is greater, since 4.5:2 is greater than 1.66:3

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Um- help-.. please don’t comment if you don’t know

IRISSAK [1]

Answer:

Each point does NOT ALWAYS give the same amount of tickets.

11 point would most likely give 90 tickets.

Step-by-step explanation:

2 points give 18 tickets = 9 tickets per point

3 points give 27 tickets = 9 tickets per point

9 points give 72 tickets = 8 tickets per point

You can add the 2 point value with the 9 point values to find out many point it takes to get to 90 tickets.

18(2 tickets) + 72(9 tickets) = 90(11 tickets)

3 0

3 years ago

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24*43x22*8x67x890x10*3x6*1x8*1

lukranit [14]

Answer:

1032x176x67x890x30x6x8

1032×176×67×890×30×6×8x^6

1.56^13×x^6

4 0

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.

7 0

1 year ago

Letter code into<br> the answer<br> box. All CAPS,<br> no spaces.

nlexa [21]

Answer:

Try CEGI

Step-by-step explanation:

if it's wrong im sorry-

6 0

3 years ago

Read 2 more answers

I Need Assistance With This:

Tamiku [17]

Answer:

22.5 in^2

Step-by-step explanation:

h = 3 in

base 1 = 6 in

base 2 = 9 in

So plug in

A = 1/2 (3) (6 + 9)

A = 3/2 (15)

A = 22.5 in^2

5 0

3 years ago