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Darina [25.2K]
3 years ago
13

Solve please This is Composite Figures :)

Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0
The area is that of two 20 yd squares and one 20 yd circle.
.. A = 2*(20 yd)^2 +(π/4)*(20 yd)^2
.. = (2 +π/4)*(400 yd^2)
.. = (800 +100π) yd^2
.. ≈ 1114.16 yd^2

The perimeter is that of a 20 yd circle and 80 yd more.
.. P = π*20 yd + 80 yd
.. ≈ 142.83 yd
You might be interested in
How do I write 4.110 in words
myrzilka [38]
Four and eleven hundreths
6 0
4 years ago
Read 2 more answers
Write the ratio in fractional notation in lowest terms. 28 inches to 42 inches
ivolga24 [154]

28 to 42 means 28/42

We now reduce 28/42 to lowest terms.

28 ÷ 7 = 4

42 ÷ 7 = 6

We now have 4/6.

We now reduce 4/6.

4 ÷ 2 = 2

6 ÷ 2 = 3

Final answer: 2/3

7 0
3 years ago
I need help for the solution​
Sonja [21]

Answer:

\boxed{ \ dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t\ }

Step-by-step explanation:

it is a long time I have not applied Ito's lemma

I would say the following

for f(x)=x^2

f'(x)=2x

f''(x)=2

so using Ito's lemma we can write that

dY_t=2V_tdV_t+\phi^2dt

dY_t=2(\theta+\psi V_t^2)dt+2\phi V_tdW_t+\phi^2dt

dY_t=(2\theta+2\psi V_t^2+\phi^2)dt+2\phi V_tdW_t

so it comes

dY_t=(2\theta+2\psi Y_t+\phi^2)dt+2\phi \sqrt{Y_t}dW_t

3 0
4 years ago
Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviati
Ilia_Sergeevich [38]

Answer:

a) \bf 0.3446^{25}=2.7095*10^{-12}

b) 3,624.25 pounds

Step-by-step explanation:

a)

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

<em>In Excel and OpenOffice Calc, this value is found with the formula </em>

<em>=1-NORMDIST(172;160;30;1) </em>

<em> (NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172) </em>

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

\bf 0.3446^{25}=2.7095*10^{-12}

b)

Similarly, we must find a weight w such that if p is the area to the left of w, then  

\bf p^{25}>0.0001  

so  

\bf p=\sqrt[25]{0.0001}=0.6918

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

<em>In Excel and OpenOffice Calc this is found with </em>

<em>=NORMINV(1-0.6918;160;30) </em>

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds

6 0
4 years ago
What is the value of x in the equation log(x+2) - log(x-1) = log 2​
pentagon [3]

Step-by-step explanation:

log(x+2)-log(x-1)=log2

log(x+2)/(x-1)=log2

=> (x+2)=2(x-1)

=>x+2 =2x-2

=> x=4

5 0
3 years ago
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