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3 years ago

Explain the relationship between an equation in exponential form and the equivalent equation in logarithmic form.

1 answer:

8 0

$\bf 
log_{{ a}}{{ b}}=y \iff {{ a}}^y={{ b}}\qquad\qquad 
% exponential notation 2nd form
{{ a}}^y={{ b}}\iff log_{{ a}}{{ b}}=y\\
\\ \quad \\
thus\\
----------------------------\\
8^5=32768\qquad \iff\qquad log_8(32768)=5$

so... notice the example there, with 8
so.. you tell us, what is the relationship then?

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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).

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

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

Find the length of the hypotenuse of a right triangle whose legs are 5 and √2.

Blizzard [7]

Answer:

$\sqrt{27}$

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let h represent the hypotenuse, then

h² = 5² + ( $\sqrt{2}$ )² = 25 + 2 = 27 ( take the square root of both sides )

h = $\sqrt{27}$ ← exact value

6 0

3 years ago

Read 2 more answers

a third grader has a goal of solving 100 math facts in 5 minutes (or 300 seconds). how many seconds can he spend on each math fa

arsen [322]

Answer:

1 fact per 3 seconds because you divide 300 by 100

5 0

3 years ago

Read 2 more answers

Find the value of x. x x = [?] Enter

liberstina [14]

Answer:

x = 14

Step-by-step explanation:

From the given figuer...

x = 7 + 7
x = 14

Therefore, the value of x is 14.

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8 0

2 years ago

PLEASE!!!!!!!!!! HELP NEEDED QUICK

Triss [41]

We have to find the expected value for the PlayBall lottery.
The price of the ticket = $1
Prize amount = $250
If a player wins, he will be winning $249 as the price is not paid back along with the prize amount. He is spending $1, getting back $250, so the net amount he is getting back is $249.

Now we have to find the probability of winning and losing.

Number of letters from A to T = 20
Number of digits from 0 to 9 = 10

Probability of picking up the same letter that was picked on that day = 1/20
Probability of picking up the same number that was picked on that day = 1/10

Thus, the Probability of picking up the same letter and same number that was picked on that day = $\frac{1}{20}* \frac{1}{10}= \frac{1}{200}$

Thus, the probability of winning = 1/200

The probability of losing = $1- \frac{1}{200}=\frac{199}{200}$

The expected value E for the PlayBall lottery will be:

$E= \frac{1}{200}(249)- \frac{199}{200}(1) \\ \\ 
E= \frac{1}{4} \\ \\ 
E=0.25$

Thus, the option C gives the correct answer

4 0

3 years ago