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3 years ago

13

At the middle school graduation dance,the DJ played 12 slow dances,which was equal to the quotient of the number of fast dances

and 2

1 answer:

Fiesta28 [93]3 years ago

4 0

Say x is the nb of fast dances
You can deduce that 12=x/2
So x=2*12
x=24

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9a + 8 - 20-3-50<br> Answer

dem82 [27]

Answer:

9a-65

Step-by-step explanation:

simply write 9a+(add the remaining numbers all together)

5 0

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

3 years ago

1. Leo made a container to store his camping gear. The container is in the shape of a rectangular prism. The container has mater

sladkih [1.3K]

You have to multiply 3 times 2= 6 + 4 = 10 so the answer is 10 I hope this help...

7 0

3 years ago

Pls help plsplsplsplsplsplspspslpssplsplsplspslpslpslsplspsps

Ainat [17]

Answer:

<em>Option A, Option C, Option E</em>

Step-by-step explanation:

The area of a kite can be identified through 1 / 2 of the multiplication of each diagonal. The first diagonal is equivalent to the addition of 50 cm and 10 cm, while the second is of length 20 cm + 20 cm.

$1 / 2 * ( 50 + 10 ) * ( 20 + 20 ),\\1 / 2 * ( 60 ) * ( 40 ),\\( 30 ) * ( 40 ),\\\\Area of Kite = 1200 square cm$

Consider the first option. If we take a look at the bit 2 * ( 1 / 2 * 20 * 50 ) the 2 and 1 /2 cancel each other out, leaving you with 20 * 50 = 1000 . Respectively in the expression 2 * ( 1 / 2 * 20 * 10 ) the 2 and 1 / 2 cancel each other out, leaving you with 20 * 10 = 200;

$1000 + 200 = 1200 square cm,\\Option A - Correct!$

The second option considers the expression 2 * ( 1 / 2 * 20 * 60 ). Again, the 2 and 1 / 2 cancel each other out, leaving you with 20 * 60;

$20 * 60 = 1200 square cm,\\Option B - Correct!$

Option C is a similar version of option a, besides the fact that the 1 / 2 doesn't exist. Thus, Option C is incorrect!

Option D is a similar version of option a as well, but as the 2 doesn't exist, it is incorrect!

This last option, option E is taken as 1 / 2 of the multiplication of the diagonals, and thus is correct!

$1 / 2 * ( 40 * 60 ),\\1 / 2 * ( 2400 ),\\1200 square cm\\\\Correct!$

4 0

3 years ago

Solve for points. identify the vertex and indicate whether it is max or min.

Keith_Richards [23]

Answer:

min at (3, 0 )

Step-by-step explanation:

given a quadratic in standard form

y = ax² + bx + c ( a ≠ 0 )

then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

y = (x - 3)² = x² - 6x + 9 ← in standard form

with a = 1 and b = - 6 , then

$x_{vertex}$ = - $\frac{-6}{2}$ = 3

substitute x = 3 into the equation for corresponding value of y

y = (3 - 3)² = 0² = 0

vertex = (3, 0 )

• if a > 0 then vertex is minimum

• if a < 0 then vertex is maximum

here a = 1 > 0 then (3, 0 ) is a minimum

4 0

2 years ago