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poizon [28]
3 years ago
6

The range of values for the sample mean that fall within 1 standard error of the mean in a sampling distribution is M39 to M43.

What Is the mean of the sampling distribution, and what is the standard error
Mathematics
1 answer:
marusya05 [52]3 years ago
5 0

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

The range of values for the sample mean that fall within 1 standard error of the mean in a sampling distribution is M = 39 to M = 43. What Is the mean of the sampling distribution, and what is the standard error

a) mean = 41, standard error = 2

b) mean = 39 or 43, standard error = 2

c) mean = 41, standard error = 4

e) the mean and standard error are unknowns

Given Information:  

Standard error = 1

Range of values = 39 to 43

Required Information:  

Mean and standard error = ?

Answer:

Mean is 41 and Standard error is 2

Explanation:

We know that the mean and standard error are related as

s = μ + (1)ο

s = μ - (1)ο

43 = μ + (1)ο

39 = μ - (1)ο

option (a) satisfies the above equations

43 = 41 + (1)2

43 = 43

39 = 41 - (1)2

39 = 39

Therefore, the correct option is (a)

Mean is 41 and standard error is 2

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Max w = 5y1 + 3y2<br> s. a.<br> y1 + y2 ≤ 50<br> 2y1 + 3y2 ≤ 60<br> y1<br> , y2 ≥ 0
CaHeK987 [17]

The maximum value of the objective function is 330

<h3>How to maximize the objective function?</h3>

The given parameters are:

Max w = 5y₁ + 3y₂

Subject to

y₁ + y₂ ≤ 50

2y₁ + 3y₂ ≤ 60

y₁ , y₂ ≥ 0

Start by plotting the graph of the constraints (see attachment)

From the attached graph, we have:

(y₁ , y₂) = (90, -40)

Substitute (y₁ , y₂) = (90, -40) in w = 5y₁ + 3y₂

w = 5 * 90 - 3 * 40

Evaluate

w = 330

Hence, the maximum value of the function is 330

Read more about objective functions at:

brainly.com/question/26036780

#SPJ1

7 0
1 year ago
A spool of rope contains 67 yards of rope. How many​ 4-foot pieces of rope can be cut from the​ spool? How much rope is left​ ov
ehidna [41]

Answer:

50 pieces with 3 inches left

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67*3 = 201

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6 0
3 years ago
onsider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutti
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Answer:

V (max)  = 2 ft³

and x side of the base is  x  =  0,5 feet

Step-by-step explanation:  See annex ( two different cubes)

We have a square piece of cardboard of  3 inches wide

Let  x be lenght of side to cut in each corner

Then the base of the (future cube) is    3  -  2x,  and the area is  

( 3 - 2x )²

And  the height   is x  Then volume of the cube as a function of x is:

V(x)  =  ( 3 - 2x )² *x   or       V(x)  =    ( 9 + 4x² - 12x )*x

V(x)  =  4x³  -  12x²  + 9x

Taking derivatives on both sides of the equation

V´(x)  =  12x² - 24x  + 9

V´(x)  =  0       12x² - 24x  + 9  =  0   simplifying   4x² - 8x  + 3  =  0

Second degree equation solving for x

x₁,₂ =  [ 24 ± √( 576) - 432  /24

x₁,₂ =  [24 ±√144 ]/24

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Of these two values we have to dismiss x₁  because if  x = 1.5 we don´t have a cube ( 0 height )

Then we take x  = 0,5  feet

And

V (max)  =  (2)²*0,5   =  4*0,5

V (max)  = 2 ft³

5 0
3 years ago
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