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Marina CMI [18]

3 years ago

15

Use the graph below for this question: graph of parabola going through negative 2, 1 and 0, 1. What is the average rate of chang

e from x = −2 to x = 0?
1
0
−2
2

1 answer:

tia_tia [17]3 years ago

6 0

Rate of change = (y2 - y1) / (x2 - x1)
r = (1 - 1) / (0 - 2)
r = 0 / -2
r = 0

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Slope intercept of 3x-2y=-16

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the slope intercept is 3/2 if you want the transfered equation y=3/2x+8

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A jacket costs 50$ at the store Evan has a coupon for 15$ off.Max pays only 0.6 of the price because his father works at the sto

Arturiano [62]

Answer:

Step-by-step explanation:

The original price of the jacket at the store is $50.

Evan has a coupon for 15$ off. This means that the amount that Evans will pay is the original price - 15% of the original price. It becomes

50 - (15/100 × 50) = 50 - 7.5 = $42.5

Max pays only 0.6 of the price because his father works at the store.

This means that the amount that Max will pay is 0.6 × the original price. It becomes

0.6 × 50 = $30

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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

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

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3 years ago

A building company bids on two large projects. The CEO believes the chance of winning the 1st is 0.6, the chance of winning the

Ksju [112]

Answer:

0.8

Step-by-step explanation:

A: P(1) = .6, P(2) = .5, P(1 and 2) = .3

P(1 or 2) = P(1) + P(2) - P(1 and 2) .6 + .5 - .3 = 0.8

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2 years ago

A diagram of a swimming pool is below. The width of the pool is 25 m, find the length of the actual pool?

lara [203]

Answer:50

Step-by-step explanation:

7.5 x 2 = 15

25 x 2 = 50

8 0

3 years ago

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