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irga5000 [103]
3 years ago
9

What is 8-6 ??????????????(((?

Mathematics
2 answers:
Tomtit [17]3 years ago
5 0

Answer:

2

Step-by-step explanation:

tatuchka [14]3 years ago
4 0

Answer: 2

:) Have a great day!

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Mark can remember only the first 4 digits of his friend's phone number. He also knows that the number has 7 digits and that the
denis-greek [22]

Answer:

<em>19800 seconds, or 330 minutes, or 5 hours + 30 minutes</em>

Step-by-step explanation:

<u>Number Permutations</u>

We know the phone number has 7 digits, 4 of which are known by Mark. This leaves him 3 digits to guess with. We also know the last one is not zero. The number can be represented as

XXY

Where X can be any digit from 0 to 9 and Y can be any digit from 1 to 9. The first two can be combined in 10x10 ways, and the last one can be of 9 ways, this gives us 10x10x9 = 900 possible permutations.

If each possible number takes him 22 seconds, every possibility will need

22x900=19800 seconds, or 330 minutes, or 5 hours + 30 minutes

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3 years ago
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Its <span>If Kevin and Amanda continue to train until week 16, what will their times be? 6. Do you believe a linear model best represents the relationship of the time of the runners and the weeks that passed?(Hint: look at question 5). What do you think this says about problems in the real world? Justify your thoughts in 3-4 sentences. </span> cause they are talking about minutes and per miles
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4 years ago
A certain delivery service offers both express and standard delivery. 75% of parcels are send by standard delivery and 25% are s
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Answer:

pictures are in order

Step-by-step explanation:

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3 years ago
The base of a rectangular prism has an area of 888 square meters. The height of the rectangular prism is 555 meters.
andriy [413]
If you want to find the volume, multiply the area of the base by the height. In this problem it would be 888*555, which is equal to 492,840
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3 years ago
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
4 years ago
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