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Rashid [163]
3 years ago
8

What's (8 + 3i)(3 + 5i).

Mathematics
1 answer:
gtnhenbr [62]3 years ago
7 0
Hello there!

(8 + 3i)(3 + 5i)

We must distribute each number/variable to the others in the opposite parenthesis.

8(3 + 5i) = 24 + 40i
3i(3 + 5i) = 9i + 15i^{2}

Combine like-terms.
40i + 9i = 49i.

Now, put your simplified answer in Standard Form;
15i^{2} + 49i + 24

This is your answer.

I hope this helps!
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The tables below show two functions. For each set of data:
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2 years ago
billy is creating a circular garden divided into 8 equal sections. The diameter of the garden is 12 feet. what is the area, in s
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Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo
Shkiper50 [21]

Answer:

a) P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335

b) P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452

c) P(X \geq 8) = 1-P(X

d) P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that X \sim Poisson(\lambda=4)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda=4  , Var(X)=\lambda=2, Sd(X)=2

a. Compute both P(X≤4) and P(X<4).

P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183

P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733

P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465

P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646

P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)

P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692

b. Compute P(4≤X≤ 8).

P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595

P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298

P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452

c. Compute P(8≤ X).

P(X \geq 8) = 1-P(X

P(X \geq 8) = 1-P(X

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)

P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954

P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563

P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042

P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559

4 0
3 years ago
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