Can u please answer this?

Gale wants to compare the cost of the events.

Lera25 [3.4K]

#a

The table represents all possibilities.

Yes its possible .
The table may represent Aquarium fee more or least etc.

#b

The carnival always costs the least.

No its possible but not always .
Aquarium fee is 14.50 on the other hand carnival fee starts at 15.55.

#c

The aquarium always costs the move.

Impossiible

8 0

2 years ago

0.1067 as a fraction

aev [14]

Answer:

1067/10000

Step-by-step explanation:

simple

4 0

2 years ago

PLZZ HELP 20 POINTS!!

katen-ka-za [31]

Answer:

son 60 puntos ...!!!!!!!!

8 0

2 years ago

Read 2 more answers

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

*PLEASE HELP URGENT NEED TO TURN IN TOMORROW* WILL GIVE BRAINY AND 60 OR MORE POINTS

pogonyaev

Answer:

2. 92 ounces of Sprite

3. 1/6 do not want mustard on their ham sandwiches

4. 6 hours playing volleyball

5. 32 portions

Step-by-step explanation:

2. Sam has 8 cans of Sprite. Each can is 11 2/4 ounces

We can simplify 2/4 to 1/2

Multiply the number of cans by the ounces per can

8 * 11 1/2

8 * 11.5

92 ounces

3. 1/2 of the students want ham. 1/3 of those do not want mustard

We multiply 1/2*1/3 = 1/6

1/6 of the students do not want mustard on their ham sandwiches

4. 8 hours at the beach. 3/4 of the time was spent playing volleyball

Multiply this together

8*3/4 = 8/4*3 = 2*3 = 6 hours

5. We have 4 lbs of wings. We divide it into 1/8 lbs parts

4 ÷ 1/8

Copy dot flip

4 * 8/1

32 portions

8 0

3 years ago

Can u please answer this?​

Can u please answer this?