I need help on this question can anyone explain to me the answer?

Mathematics

1 answer:

raketka [301]2 years ago

4 0

Answer:

Option C, $y = \frac{1}{10} x+1$

Step-by-step explanation:

We can find the correct answer by plugging in the x and y values of a point on the graph into each option, then finding the one that gives a true statement as a result.

For example, one point on the graph is (30,4). We can substitute 30 for the x and 4 for the y in option C's equation, $y = \frac{1}{10} x+1$ . Then, simplify:

$y = \frac{1}{10} x+1\\4 =\frac{1}{10} (30)+1\\4 = \frac{30}{10} +1\\4 = 3 +1\\4 = 4$

4 does equal 4, so when we substitute a point from the graph into this equation, the result is a true statement. Thus, option C is the answer.

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A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c

slava [35]

Answer:

$I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

$\frac{dI}{dt}=a*I*(1000-I)$

Rearranging and integrating

$\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t$

At the initial breakout (t=0) there was one trooper infected (I=1)

$-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755$

In two days (t=2) there were 5 troopers infected

$-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725$

Rearranging, we can model the number of infected troops (I) as

$-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755} \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}$