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Elena-2011 [213]
3 years ago
9

Help Please! 5=(2m-b)/(m+b)

Mathematics
2 answers:
finlep [7]3 years ago
6 0

Answer:

B= -1/2m

Step-by-step explanation:

Ion know wat your solving for buh theres b

antiseptic1488 [7]3 years ago
6 0

Answer:

m=−2b

b=−m/2

work: i just know

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At Pike Place Fish Market in Seattle, customers can purchase a variety of different types of seafood. One type of seafood sold a
allsm [11]

Answer:

z=1.28

And if we solve for a we got

a=3.6 +1.28*0.8=4.624

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

 C. 4.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(3.6,0.8)  

Where \mu=3.6 and \sigma=0.8

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.1   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.28

And if we solve for a we got

a=3.6 +1.28*0.8=4.624

So the value of height that separates the bottom 90% of data from the top 10% is 4.624.

So then the best answer for this case would be:

 C. 4.64

8 0
3 years ago
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