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3 years ago

12

Zach has a box with a volume of 48 cubic units. Which could be the dimensions of Zach's box?

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Step-by-step explanation:

<em>It will be A And C.</em>

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Evaluate the expression

lora16 [44]

The correct answer is B=4

5 0

3 years ago

What is the simplified form of this expression?<br><br> 7a2 + 10a − 3a − 2a2

nikklg [1K]

Answer:

What is the simplified form of this expression?

7a2 + 10a − 3a − 2a2

Step-by-step explanation:

7a2 + 10a − 3a − 2a2

Combine like terms

742 + 10a - 3a - 2a2

742 + 7a - 2a2

solution=

5a2 + 7a

6 0

3 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).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

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

Solve for x.<br> Need help, please.

dlinn [17]

x=9

hope this helps you. :)

5 0

3 years ago

What is the slope if the line through the points 2/3, 4/7 and 2/3, 11/7

Umnica [9.8K]

The fromula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points $\left(\dfrac{2}{3},\ \dfrac{4}{7}\right)$ and $\left(\dfrac{2}{3},\ \dfrac{11}{7}\right)$

Substitute:

$m=\dfrac{\frac{11}{7}-\frac{4}{7}}{\frac{2}{3}-\frac{2}{3}}=\dfrac{\frac{7}{7}}{0}\ !!!!$

<h3>Answer: slope is undefined.</h3><h3 />

It's a vertical line: $x=\dfrac{2}{3}$

6 0

3 years ago