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3 years ago

12

A company produces optical-fiber cable with a mean of 0.6 flaws per 100 feet. What is the probability that there will be exactly

4 flaws in 1000 feet of cable

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

$P(X = 4) =0.133$

Step-by-step explanation:

From the question we are told that:

Mean $\x=0.6/100$

Flaws $f=4$

Distance $d_2=1000ft$

Generally the equation for Poisson mean lambda over 1000 is mathematically given by

$\lambda=0.6*1000/100$

$\lambda = 6$

Therefore

$P(X = 4) = {e^-\lambda * \lambda^{\=x}} {{\=x}!}$

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

$P(X = 4) =0.133$

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Answer:

Step-by-step explanation:

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3 years ago

What is -42.15+12.45

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3 years ago

Read 2 more answers

The results of a mathematics placement exam at two different campuses of Mercy College follow: Campus Sample Size Sample Mean Po

Leona [35]

Answer:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

$p_v =P(Z>3.37)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.

Step-by-step explanation:

Data given

Campus Sample size Mean Population deviation

1 330 33 8

2 310 31 7

$\bar X_{1}=33$ represent the mean for sample 1

$\bar X_{2}=31$ represent the mean for sample 2

$\sigma_{1}=8$ represent the population standard deviation for 1

$\sigma_{2}=7$ represent the population standard deviation for 2

$n_{1}=330$ sample size for the group 1

$n_{2}=310$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

P value

Since is a one right tailed test the p value would be:

$p_v =P(Z>3.37)=1-P(Z$

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the mean for the Campus 1 is significantly higher than the mean for the group 2.

5 0

3 years ago

Can someone work this problem out for me? 1/3 - 1/6 +4/18. please and thank you.

NISA [10]

Simplify the following:
1/3 - 1/6 + 4/18

The gcd of 4 and 18 is 2, so 4/18 = (2×2)/(2×9) = 2/2×2/9 = 2/9:
1/3 - 1/6 + 2/9

Put 1/3 - 1/6 + 2/9 over the common denominator 18. 1/3 - 1/6 + 2/9 = 6/18 - 3/18 + (2×2)/18:
6/18 - 3/18 + (2×2)/18

2×2 = 4:
6/18 - 3/18 + 4/18

6/18 - 3/18 + 4/18 = (6 - 3 + 4)/18:
(6 - 3 + 4)/18

6 + 4 = 10:
(10 - 3)/18

10 - 3 = 7:

Answer: 7/18

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4 years ago

Randy is walking home from school. According to the diagram above, what is his total distance from school to home? Show your wor

In-s [12.5K]

Answer:

First, when he walks, we can see in the image that between the school and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Then he needs to walk 2km.

Now if he has a jet-pack, he can ignore the buildings and just take the shorter path, here we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the catheti is the vertical distance (two blocks of 0.5km, so this catheti has a length of 2*0.5km = 1km), and the other one is the horizontal distance, also 1km.

The actual distance of this path is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, in this case the distance and the displacement would be the same.

This is because the definitions of distance and displacement are:

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km and the displacement is 1.41km , but when he uses the jet pack, the distance is equal to the displacement, both are 1.41km.

7 0

4 years ago