What do you call a characteristic of a living thing that helps it survive?

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

What is the value of 2 in 255.6

True [87]

Answer:

200?

Step-by-step explanation:

its kinda simple but the 2 is in the hundreds place therefore it is 200.

hope it helped ;)

6 0

4 years ago

The authors of a paper presented a correlation analysis to investigate the relationship between maximal lactate level x and musc

ArbitrLikvidat [17]

Answer:

a) Sample correlation coefficient, r = 0.7411

bi) test statistic, t = 4.102

bii) P-value = 0.000736

Step-by-step explanation:

a) The formula for the sample correlation coefficient is given by the formula:

$r = \frac{S_{xy} }{\sqrt{S_{xx} S_{yy} }} }$

$S_{xx} = 2,648,130.357\\S_{yy} = 36.7376,\\S_{xy} = 7408.225$

$r = \frac{7408.225}{\sqrt{2648130.357*36.7376} }$

r = 0.7511

b)

i) formula for the test statistic is given by the formula:

$t = \frac{r\sqrt{n-1} }{\sqrt{1 - r^{2} } }$

sample size, n = 4

$t = \frac{0.7511\sqrt{14-1} }{\sqrt{1 - 0.7511^{2} } }$

t = 4.102

ii) Degree of freedom, df = n -2

df = 14 -2

df = 12

The P-value is calculate from the degree of freedom and the test statistic using excel

P-value =(=TDIST(t,df,tail))

P-value = (=TDIST(4.1,12,1)

P-value = 0.000736

4 0

3 years ago

Please try to explain step by step<br>Thanks in advance

guajiro [1.7K]

Answer:

you cant answer it without a calculator

Step-by-step explanation:

press shift - then cos then whatever u need to do to get a

5 0

3 years ago

Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph

kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0 and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of f(y) versus y

Looking at the given differential equation

$\frac{dy}{dt} = e^{y} - 1$ for -∞ < $y_{o}$ < ∞

We can say let $\frac{dy}{dt}$ = f(y) = $e^{y} - 1$

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

To fin the critical point we have to set $\frac{dy}{dt}$ = 0

This means $e^{y} - 1$ = 0

For this to be possible $e^{y}$ = 1

which means that $e^{y}$ = $e^{0}$

which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0 hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where f(y) = 0.

In each of the intervals bounded by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

5 0

3 years ago