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baherus [9]
3 years ago
7

If initial population of any place is Po,

Mathematics
1 answer:
g100num [7]3 years ago
7 0

Answer:

please join my meeting

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Por favor ayúdeme con mi tarea de matemáticas... Son ecuaciones<br> 2x + 4y=10<br> -x+5=9
Hatshy [7]
X+2y-5=0
x=-4



-x=9-5
-x=4
x=-4
3 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
A random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength
irinina [24]

Answer:

Probability that the sample mean comprehensive strength exceeds 4985 psi is 0.99999.

Step-by-step explanation:

We are given that a random sample of n = 9 structural elements is tested for comprehensive strength. We know the true mean comprehensive strength μ = 5500 psi and the standard deviation is σ = 100 psi.

<u><em>Let </em></u>\bar X<u><em> = sample mean comprehensive strength</em></u>

The z-score probability distribution for sample mean is given by;

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean comprehensive strength = 5500 psi

            \sigma = standard deviation = 100 psi

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample mean comprehensive strength exceeds 4985 psi is given by = P(\bar X > 4985 psi)

    P(\bar X > 4985 psi) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{4985-5500}{\frac{100}{\sqrt{9} } } ) = P(Z > -15.45) = P(Z < 15.45)

                                                                  = <u>0.99999</u>

<em>Since in the z table the highest critical value of x for which a probability area is given is x = 4.40 which is 0.99999, so we assume that our required probability will be equal to 0.99999.</em>

4 0
3 years ago
You spend 2 hours and 15 minutes commuting to work each day. You work five days per week. How much time do you spend commuting t
xeze [42]

11 hours and 15 minutes because doing 2.25, which is equal to 2 hours and 15 minutes, times 5 you end up getting 11.25, and converting that to time you get 11 hours and 15 minutes.

8 0
3 years ago
Ron practiced piano 1 2/3 hours on monday and 5/6 hour on tuesday. how much did he practice in all those two days
insens350 [35]
2.5 hours i think but i might be wrong :)
4 0
3 years ago
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