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3 years ago

This is the last questionn help

Mathematics

1 answer:

makkiz [27]3 years ago

7 0

Answer:

Step-by-step explanation:

That's the answer....

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If P(E) = 0.32, then what are the odds against E?

steposvetlana [31]

Answer:

0.68 : 0.32

Step-by-step explanation:

the probability of all the outcomes of an event is always 1 .

if P(E) = 0.32,

P(E') = 1 - 0.32

= 0.68

N.B that P(E') is the same as P( not E ).

since your answer should be given in the ratio form, as the odds against E ( probability not E : probability E),

P(E') : P(E)

0.68 : 0.32

hope this helps you!

-s.

7 0

2 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

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

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago

What is 14/10 in percent form?

Lilit [14]

Get the value of the decimal:
$\frac{14}{10}=1.40$

Remember that percentages are simply the value of a decimal multiplied by 100. (From the Latin, "per centus", meaning "per 100")

therefore:
$1.40(100)=140percent$