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3 years ago

Please guys help to solve this problem

Mathematics

2 answers:

liq [111]3 years ago

6 0

9514 1404 393

Answer:

300

Step-by-step explanation:

Since nobody failed, the number who passed one or the other was 100%.

P(O + W) = P(O) +P(W) -P(O&W)

100% = 80% +70% -P(O&W)

P(O&W) = 50% = 150 students

If 150 students are 50% of the examinees, then 100% will be 300 students.

zhuklara [117]3 years ago

3 0

Answer:

hope it helps

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A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

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Ulleksa [173]

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trapecia [35]

Answer:

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We are given that Alice and Bob are playing a game .

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