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natulia [17]
1 year ago
9

The blade length is the radius of the circle formed by the area swept by the blades to the nearest square foot, what is the area

of this circle for model 3
Mathematics
2 answers:
USPshnik [31]1 year ago
8 0

Answer:The blade length is the radius of the circle formed by the area swept by the blades. To the nearest square foot, what is the area of this circle for model 3?

Step-by-step explanation:

rewona [7]1 year ago
3 0
Sorry i do not know wish i did 
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How can the triangles be proven similar by the SSS similarity theorem? Show that the ratios StartFraction U V Over X Y EndFracti
dangina [55]

Answer:

<u>option 1</u>

Step-by-step explanation:

The question is as shown in the attached figure.

From the figure, we can deduce that:

∠U ≅ ∠X and ∠W ≅ ∠Z  and ∠V ≅ ∠Y

So, ΔUWV similar to ΔXZY

But it is required to know " How can the triangles be proven similar by the SSS similarity theorem? "

So, we need to prove the corresponding angles are proportion

With the help of the previous corresponding angles.

So, we need to show that the ratios \frac{UV}{XY}  \ , \ \frac{WU}{ZX} \  and \ \frac{WV}{ZY} are equivalent.

So, the answer is <u>option 1</u>

5 0
1 year ago
Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ
SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

\ell_i=\dfrac{i-1}2

r_i=\dfrac i2

with 1\le i\le10.

These subintervals have midpoints given by

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

Over each subinterval, we approximate f(x) with the quadratic polynomial

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that the integral we want to find can be estimated as

\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that

\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6

so that the arc length is approximately

\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

5 0
1 year ago
The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution
oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!

Then, we can calculate P(X ≤ 5) as:

P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080

4 0
1 year ago
The bird house in your friend's yard casts a shadow that is 14 feet long. Your friend is 5 feet tall and casts a shadow that is
astra-53 [7]
3.5/5=14/x   20ft tall
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1 year ago
Marianne has been collecting donations for her biscuit stall at the school summer fayre. There are some luxury gift tins of bisc
Marta_Voda [28]
I don't know what 10p is so before I answer I'd have to know what that is
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2 years ago
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