X

If a person ran 1/2 of a mile in 1/10 of an hour. How far will the person run in one (1) hour?

makvit [3.9K]

Step-by-step explanation:

In 1/10 of 1hr = 1/2 mile

10*(1/10) of 1hr = 10(1/2) mile

(10/10) of 1hr = (10/2) mile

1 of 1hr = 5mile

It says, he can run 5 miles in 1 hour

5 0

3 years ago

Four girls have 2.20 to share equally. How much money will each girl get?

Nata [24]

Answer:

0,55

Step-by-step explanation:

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

Angle problem -math

Jet001 [13]

Its D because vertical angles are right across from each other!

6 0

2 years ago

Which terms could be used as the first term of the expression below to create a polynomial written in standard form? Select five

Alina [70]

Answer:

3r⁴s⁵

- r⁴s⁶

- 6rs⁵

Step-by-step explanation:

1) The expression given is:

+ 8r²s⁴ – 3r³s³

2) The choices given are:

s⁵

3r⁴s⁵

- r⁴s⁶

- 6rs⁵

1) A polynomial written in standard form has the terms ordered in decreasing order.

For example, 4x⁵ + 8x⁴ + 3x² + 27

2) The given polynomial has degree 6: + 8r²s⁴ – 3r³s³, and it can be ordered respect any of the two variables either r or s.

3) Adding a new term as first term needs to have degree equal or greater than 6.

So, the candidates from the list are: 3r⁴s⁵, - r⁴s⁶,- 6rs⁵

4) If you use 3r⁴s⁵, the polynomial in standard form would be:

3r⁴s⁵ – 3r³s³ + 8r²s⁴ (as you see the degree of r is decreasing from left to right).

5) If you use - r⁴s⁶, the polynomial in standard form would be:

- r⁴s⁶ – 3r³s³ + 8r²s⁴ (as you see, the degree of r decreases from left to right)

6) If you use - 6rs⁵, the polynomial in standard form would be

- 6 rs⁵– 3r³s³ + 8r²s⁴ (as you see the degree of s is decreasing from left to right).

7) You cannot use s⁵ as first term because its degree is less than 6.

3 0

3 years ago

Read 2 more answers