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Novosadov [1.4K]

3 years ago

10

Find two numbers the quotiens is between 136÷6

1 answer:

Alekssandra [29.7K]3 years ago

6 0

The answer is 22.6 so the quotient would be between 22.0 and 23.0, hope this helps.

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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

If c is a positive integer, how does the graph of y=x+c compare to the graph of y=x?

Len [333]

Answer:

It is shifted c units to the left .

Step-by-step explanation:

In general, adding a constant c to the value of x shifts a graph c units to the left.

The graph of y = x + c is parallel to y = x, the y-intercept becomes (0, c) and the x-intercept becomes (-c, 0).

In the diagram below, the red line is the graph of y = x, and the blue line is the graph of y = x + 1.

3 0

3 years ago

Cost of a New Car From 1970 to 1990, the average cost of a new car, C

Lunna [17]

Answer:

In year 1986 the average cost of car is $12,000

Step-by-step explanation:

Cost function:

C = 30.5t² + 4192

Cost = $12,000

Find:

Year

Computation:

C = 30.5t² + 4192

12,000 = 30.5t² + 4192

t = 16 year

So,

Year 1970 + 16 year

1986

So, In year 1986 the average cost of car is $12,000

3 0

3 years ago

Simplify the linear expression.

Molodets [167]

Answer: The answer to the first is -29x-5. The Answer to the second is -5/6a - 5/8

Step-by-step explanation:I took the K12 test I hope this helps.

4 0

3 years ago

Indicate whether each of the following fractions is proper or improper. a. 4⁄16 b. 75⁄70 c. 2⁄15 d. 6⁄6

Westkost [7]

75/70 is an Improper Fraction

7 0

3 years ago

Read 2 more answers