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vagabundo [1.1K]

1 year ago

12

One jar filled with honey weighs 120 grams and one empty jar with half filled honey jar weighs 150 grams what is the weight of t

he jar

1 answer:

VMariaS [17]1 year ago

5 0

Answer:

90 grams

Step-by-step explanation

Full = 120

half filled + empty = 150

half of 120 = 60

Therefore, a jar weighs 150 - 60

= 90 grams

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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

△QRS is rotated about point P to △Q′R′S′ .

Sedbober [7]

Answer:

As Given △QRS is rotated about point P to △Q′R′S′ .

As we know after rotation by any angle of pre-image neither the shape nor size of the image changes, the image after rotation is congruent to pre image.

Out of the options given

→The corresponding lengths, from the point of rotation, between the pre-image and the image are preserved.

→The corresponding angle measurements in each triangle between the pre-image and the image are preserved.

are true.

4 0

2 years ago

leda bakes 73 cupcakes and accidentally drop 5 of them. she divided them equally into 13 containers for a bake sale for the scho

Vadim26 [7]

If she dropped 5 out of 73 . The answer would be 68 . Or are you asking how many are in the container

5 0

2 years ago

-129 = -48 - n <br> Thank you

Flauer [41]

−129=−48−n

Simplify:

−129=−48+−n

Flip the equation.

−n−48=−129

Add 48 to both sides.

−n−48+48=−129+48

−n=−81

Divide both sides by -1.

−n/−1 = −81/−1

n=81

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2 years ago

Determine whether each view represents the object. Assume there are no hidden cubes. HELP PLEASE I DON'T UNDERSTAND!!

zubka84 [21]

Answer:

The answer is: #3 yes, yes

Step-by-step explanation:

The first 3 shapes in the image count as the first view and the 4th shape is the second view. They both represent the object.

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2 years ago