Help please!! really quickly

To get the ratio of three green marbles and six blue marbles will there be in simplest form?

katen-ka-za [31]

Answer:

Well your question is a little unclear but say there is a bag of marbles containing 3 green marbles and 6 blue marbles then if you wanted to find the ratio of the green marbles it would be 3/9 or 1/3

8 0

3 years ago

Read 2 more answers

I need help pls help

Snezhnost [94]

Answer:

c.

Step-by-step explanation:

hope this helps

3 0

3 years ago

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Indicate an interval that is decreasing, and explain your reasoning.

leva [86]

Answer:

(-1,1) because -1 and 1 are not inclusive

Step-by-step explanation:

from x = -1 to x = 1, the y-value keeps decreasing from 4 to 2

there is another one, (3,5)

4 0

3 years ago

Marvin invested 65% of his retirement account in stocks and 35% in gold. Marvin believes that the return to stocks over the next

ser-zykov [4K]

Answer:

1)

the mean rate of return is 17 %

the standard deviation of return is 17.06055

2)

the probability that Marvin's portfolio will make at least 20% over the next 12 months is 0.4325

Step-by-step explanation:

Given the data in the question;

1)

For the portfolio, the mean return and standard deviation are computed as follows;

Mean = Return = 0.65 × 10 + 0.35 × 30

= 6.5 + 10.5

= 17 %

Therefore, the mean rate of return is 17 %

Standard deviation will be;

σp = √( 0.65² × 15² + 0.35² × 40² )

= √( 0.4225 × 225 + 0.1225 × 1600 )

= √( 95.0625 + 196 )

= √291.0625

= 17.06055

Therefore, the standard deviation of return is 17.06055

2)

probability that Marvin's portfolio will make at least 20% over the next 12 months.

P( X > 20 )

we convert to a standard normal variable;

Z = $\frac{20-17}{17.06055}$ )

Z = 0.17

from z table, p-value is;

p( X < 20 ) = 0.5675

P( X > 20 ) 1 - 0.5675 = 0.4325

Therefore, the probability that Marvin's portfolio will make at least 20% over the next 12 months is 0.4325

5 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

=0

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

=0

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago