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Please explain your reasoning behind your answers! I will give 5 stars to the person who gives accurate answers AND thorough explanations!



Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.6 and standard deviation 0.79.



at most 3.00    
between 2.6 and 3.00    

Solution

z = x – m / (sd / sqrt(n) )

a)

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ii) P(2.6 < x < 3) = P( (2.6 – 2.6) / (0.79 / sqrt(25)) < z < 2.53 ) = P(0<z<2.53)

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On May 1, 2008, the forest service stocked a lake with

On May 1, 2008, the forest service stocked a lake with 136 bluegill fish. On October 1, five months later, they estimated the population of bluegill in the lake to be 197. Assume the population of bluegill increased exponentially.

For this population of bluegill, find:

the 5-month percent change:  

the 5-month growth factor:  

the 1-month growth factor:  

the 1-month percent change:   

Define a function f that gives the number of fish in the lake t months after May 1, 2008.

Define a function g that gives the number of fish in terms of n , the number of 5-month periods after May 1, 2008.

Use a graphing calculator and the functions you defined above to determine how many months after May 1 are required for the population to surpass 750 bluegill. Your answer should be a whole number of months.

Solution

let Po be the initial population

and P is the final population

then , P = Po e^kt

in 5 months the polpulation has increased from 136 to 197

197 = 136 e^5k

on solving we get

k = .0740

percentage change of 5 months = 7.4%

growth factor is .0740

growth factor of 1 month is plug t = 1 in the equation P = Po e^kt

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Please explain the reasoning behind your answers so I c

Please explain the reasoning behind your answers so I can understand! I will rate 5 stars to the first person who gives accurate answers AND provides the best explanation.


The inside diameter of a randomly selected piston ring is a random variable with mean value 12 cm and standard deviation 0.07 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.)



Solution

Ok this is the exact (not rounded off solution)

u=12 s=0.07

a) x1=11.99 z1=(x1-u)/s z1=-0.1428

x2=12.01 z2=(x2-u)/s z2=0.1428

P(11.99<X<=12.01) = P(-0.1428<Z<=0.1428)

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A person drinks a cup of coffee with 100mg of caffeine.

A person drinks a cup of coffee with 100mg of caffeine. Caffeine has a half-life of 5.7 hours. 1. Write a formula for the amount of caffeine left in the persons body t hours later.2. Find the amount of caffeine left after 8 hours.3. When will there be less than 20mg of caffeine left in the persons body?

Solution

Gen equation : y(t) = Ae^(-kt)

A = 100 mg ; y = 100e^(-kt) ; t1/2 = 5.7 hrs

T calculate k ; 50 = 100e^(-5.7k)

0.2 = e^(-5.7k)

ln(0.2) = -5.7k

k = 0.282

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A travel association claims that the mean daily meal co

A travel association claims that the mean daily meal cost for two adults traveling together on vacation is $125. A random sample of 25 such groups of adults has a mean daily meal cost of $120 and a standard deviation of $6. is there enough evidence to reject the claim at a=.05?

B). calculate the standardized test statistic:

t= (round to 3 decimal places)

Solution

H0: mean daily meal cost for two adults traveling together on vacation =125

H1: mean    daily meal cost for two adults traveling together on vacation  not equal to 125


(standardized test statistic)=t= (120-125)/(6/sqrt 25)= -4.167


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pre calculus Solve the triangle ABC, if the triangle ex

pre calculus

Solve the triangle ABC, if the triangle exists. B=31 degree 18\’ a = 37.6 b=32.1 Select the correct choice below and fill in the answer boxes within the choice. There is only 1 possible solution for the triangle. The measurements for the remaining angles A and C and side c are as follows. m A= m C = degree \'(Simplify your answer. Round to the nearest degree as needed. Round to the nearest minute as needed.) The length of side c (Round to the nearest tenth as needed.) There are 2 possible solutions for the triangle. The measurements for the solution with the longer side c arc as follows. m A= m C = degree \'(Simplify your answer. Round to the nearest degree as needed. Round to the nearest minute as needed.) The length of side c (Round to the nearest tenth as needed.) There are 2 possible solutions for the triangle. The measurements for the solution with the shorter side c arc as follows. m A= m C = degree \'(Simplify your answer. Round to the nearest degree as needed. Round to the nearest minute as needed.) The length of side c (Round to the nearest tenth as needed.) There are no possible solutions for this triangle.

Solution

B = 31.18\’ b = 32.1 and a= 37.6

Apply sine rule :

b/sinB = a/sinA ; sinA = a*sinB/b

sinA = (37.6 *sin31.18)/32.1 = 0.608

A = 37.48 . sin is +ve in Ist and IInd quadrant.

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Graphical solutions to linear programming problems have

Graphical solutions to linear programming problems have an infinite number of possible objective function lines.

Solution

Answer: TRUE

Diff: 2             Page Ref: 34

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A (3i + 2j + k) in B (4j) in and C (i + 4j 3k) in.

A = (3i + 2j + k) in; B = (4j) in; and C = (i + 4j 3k) in.

Evaluate (A * B)*C

Solution

A = (-3i +2j +k) B = 4j C = ( -i +4j -3k)

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A signal which can be green or red with probability 45

A signal which can be green or red with probability 4/5 or 1/5 respectively,is received by station A and transmitted to station B,the probablity of each station receiving the signal coreectly is 3/4.if hte signal received at station B is green,then the probabilty of original signal was red is?

Solution

Green and correct 4/5 *3/4= 12/20

Green and incorrect 4/5 *1/4= 4/20

Red and correct 1/5 *3/4= 3/30

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