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dalvyx [7]
3 years ago
6

1) Given P(A) = 0.3 and P(B) = 0.5, do the following.

Mathematics
1 answer:
Delicious77 [7]3 years ago
6 0

Answer:

1) a) 0.8

b) 0.6

2) a) 0.08

b) 0.14

Step-by-step explanation:

1) Given

P(A) = 0.3 and P(B) = 0.5

Let us learn about a formula:

P(A\ or\ B) = P(A) +P(B) -P(A\ and\ B)\\OR\\P(A\cup B) = P(A) +P(B) -P(A\cap B)

(a) If A and B are mutually exclusive i.e. no common thing in the two events.

In other words:

P(A\ and\ B) = P(A \cap B) = 0

Using above formula:

P(A\ or\ B) = P(A) +P(B) -P(A\ and\ B)\\\Rightarrow P(A\ or\ B) = 0.3 + 0.5 -0 = \bold{0.8}

(b)  P(A and B) = 0.2

Using above formula:

P(A\ or\ B) = P(A) +P(B) -P(A\ and\ B)\\\Rightarrow P(A\ or\ B) = 0.3 + 0.5 -0.2 = \bold{0.6}

*************************************

1) Given

P(A) = 0.4 and P(B) = 0.2

Let us learn about a formula:

P(A\ and\ B) = P(B) \times P(A/B)  for dependent events

P(A\ and\ B) = P(A) \times P(B) for independent events.

(a) If A and B are independent events :

Using the above formula for independent events:

P(A\ and\ B) = 0.4 \times 0.2 = \bold{0.08}

(b)  P(A / B) = 0.7

Using above formula:

P(A\ and\ B) = P(B) \times P(A/B) = 0.2 \times 0.7 = \bold{0.14}

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Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti
Vikki [24]

Answer:

General Solution is y=x^{3}+cx^{2} and the particular solution is  y=x^{3}-\frac{1}{2}x^{2}

Step-by-step explanation:

x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}

This is a linear diffrential equation of type

\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)..................(i)

here p(x)=\frac{-2}{x}

q(x)=x^{2}

The solution of equation i is given by

y\times e^{\int p(x)dx}=\int  e^{\int p(x)dx}\times q(x)dx

we have e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}

Thus the solution becomes

\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+cy=x^{3}+cx^{2

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have 6=8+4c

Thus solving for c we get c = -1/2

Thus particular solution becomes

y=x^{3}-\frac{1}{2}x^{2}

5 0
3 years ago
Help numbers 3 pleaaaaaassssseeee
Varvara68 [4.7K]
It is 2 4/5, 2 4/5 is 2.8, 4/5 is greater than 3/4
3 0
2 years ago
Read 2 more answers
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