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SVETLANKA909090 [29]
3 years ago
13

A distribution center receives shipments from three different factories in quantities of 50, 35, and 20. Three times a product i

s selected at random. Find the probability that none of the products came from the second factory.
Mathematics
1 answer:
sladkih [1.3K]3 years ago
8 0
The problem ask to compute and calculate the probability of the said problem if the distribution center receives a shipment of 3 different factories in quantities of 50,35 and 20. So base on that quantity the possible answer would be 2/3. I hope you are satisfied with my answer and feel free to ask for more 
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The product 3/4 times 1/2 will be less than or greater than 3/4
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Step-by-step explanation:

first you would multiply and get 4/8, which simplifies to 1/2, which would be 2/4, which is less than 3/4

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Find the area of a circle with a circumference
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The weights of soy patties sold by a diner are normally distributed. A random sample of 25 patties yields a mean weight of 4.2 o
just olya [345]

Answer:

t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2

The degrees of freedom are given by:

df =n-1=25-1=24

And the p value would be given by:

p_v = P(t_{24}>2) =0.0285

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate      

Step-by-step explanation:

Data provided

\bar X=4.2 represent the sample mean for the weigths

s=0.5 represent the sample standard deviation

n=25 sample size      

\mu_o =4 represent the value that we want to analyze

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to conduct a hypothesis in order to check if the true mean weigth is less than 4 or not, the system of hypothesis would be:      

Null hypothesis:\mu \leq 4      

Alternative hypothesis:\mu > 4      

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2  

The degrees of freedom are given by:

df =n-1=25-1=24

And the p value would be given by:

p_v = P(t_{24}>2) =0.0285

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate

5 0
3 years ago
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