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Andrei [34K]
3 years ago
11

Find the area of the polygon.QUICKLY PLEASE

Mathematics
2 answers:
Serggg [28]3 years ago
8 0

Answer:

72 units squared

Step-by-step explanation:

12*12 / 2

Irina18 [472]3 years ago
4 0

Answer:

72

Step-by-step explanation:

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Which is the larger amount?<br> A) 5 cups<br> B) 500 milliliters
kvasek [131]

Answer:

a

Step-by-step explanation:

8 0
3 years ago
If the standard deviation of returns on the market is 20 percent, and the beta of a well-diversified portfolio is 1.5. Calculate
Luden [163]

Answer:

a. 30 percent.

Step-by-step explanation:

Given that:

The standard deviation of returns = 20 percent

Beta = 1.5

Beta=Standard deviation of portfolio × correlation/Standard deviation of market × Correlation

Since Correlation with the market will be +1;

Then;

The Standard deviation of portfolio = 1.5 × 20%

The Standard deviation of portfolio = 30.00%

3 0
3 years ago
Complete the table for the given rule
irinina [24]

Answer:

see explanation :)

Step-by-step explanation:

y = − x − 4

is in the slope-intercept form for a linear equation,  

y = m x + b , where  m  is the slope and  b  is the y-intercept. In the given equation,  

m = − 1  and  b = − 4 .


Ordered Pairs

x ... ... . . y

4 ... . − 8

2 ... . − 6

0 ... . − 4

− 2 . ...− 2

− 4 ... . . 0

7 0
3 years ago
Read 2 more answers
4.8.4 Raw materials are studied for contamination. Suppose that the number of particles of contamination per pound of material i
forsale [732]

Answer:

700 lbs and 3.742

Step-by-step explanation:

Given that raw materials are studied for contamination. Suppose that the number of particles of contamination per pound of material is a Poisson random variable with a mean of 0.02 particle per pound.

a) To get 14 particles of contamination we must have

expected number of pounds = \frac{14}{0.02} \\=700

b) Since for 700 pounds particles expected are 14 we can say in  a Poisson distribution mean and variance are equal

i.e. Var (700 lbs to obtain 14 particles of contamination ) = 14

Std deviation =\sqrt{14} =3.742

5 0
3 years ago
Please I need help with differential equation. Thank you
Inga [223]

1. I suppose the ODE is supposed to be

\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)

Solving for \dfrac{\mathrm dy}{\mathrm dt} gives

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}

which is undefined when t=\pm1. The interval of validity depends on what your initial value is. In this case, it's t=-\dfrac12, so the largest interval on which a solution can exist is -1\le t\le1.

2. Separating the variables gives

\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}

Integrate both sides. On the left, we have

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}

where we substituted z=y^{1/2} - or z^2=y - and 2z\,\mathrm dz=\mathrm dy - or \mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}.

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)

On the right, we have

\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)

\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C

So

2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C

\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C

y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}

y^{1/2}=C(1-t^2)^{1/4}-1

I'll leave the solution in this form for now to make solving for C easier. Given that y\left(-\dfrac12\right)=1, we get

1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1

2=C\left(\dfrac54\right)^{1/4}

C=2\left(\dfrac45\right)^{1/4}

and so our solution is

\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}

3 0
3 years ago
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