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3 years ago

Can someone help me on this Please

Mathematics

1 answer:

saul85 [17]3 years ago

6 0

9514 1404 393

Answer:

11.6 cm

Step-by-step explanation:

As the page title tells you, the Pythagorean theorem must be applied more than once. As you know, it tells you the square of the hypotenuse is the sum of the squares of the two sides.

AD² = ED² +EA²

EA² = AD²-ED² = 7² -6² = 13

EA = √13 ≈ 3.606

CD² = ED² +EC²

EC² = CD² -ED² = 10² -6² = 64

EC = √64 = 8

The length of the horizontal diagonal is ...

AC = EA +EC = 3.6 +8 = 11.6 . . . cm

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

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3 years ago

Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8

Kitty [74]

Substitute $z=\ln x$ , so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)$

Then the ODE becomes

$x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0$
$\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0$

which has the characteristic equation $r^2+1=0$ with roots at $r=\pm i$ . This means the characteristic solution for $y(z)$ is

$y_C(z)=C_1\cos z+C_2\sin z$

and in terms of $y(x)$ , this is

$y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$

From the given initial conditions, we find

$y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1$
$y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8$

so the particular solution to the IVP is

$y(x)=\cos(\ln x)+8\sin(\ln x)$

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