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Andreas93 [3]
2 years ago
15

In developing an interval estimate for a population mean, a sample of 50 observations was used. the interval estimate was 19.76

± 1.32. had the sample size been 200 instead of 50, the interval estimate would have been:
Mathematics
1 answer:
alexandr402 [8]2 years ago
4 0

 

The formula for interval estimate would be: μ = M ± Z(<span>sM</span>)

Where: μ is estimate

M is the mean 

Z is the z value

(<span>sM</span>) is the standard error

 

μ = M ± Z(<span>sM</span>)

n = 200 rather than 50 (√200 = 2√50)

<span>⇒ ME = (1/2) * 1.32 = .66</span>

<span>Using the formula above, plugging this in will give us: μ = 19.76 ± .66</span>

<span> = 19.76  ±  .66 is the confidence interval or interval estimate</span>

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qwelly [4]

The equation would be y=2.10x+3.25!


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6 0
2 years ago
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In the roof truss, line BG bisects ∠ABC and ∠DEF, m∠ABC= 112°, and ∠ABC≅∠DEF.
Dennis_Churaev [7]

In the given diagram,  line BG bisects ∠ABC and ∠DEF, m∠ABC= 112°, and ∠ABC≅∠DEF.  So the measure of angles are :

1. m∠DEF=  112° because ∠ABC≅∠DEF

2. m∠ABG=  56° because a straight line BG is bisecting ∠ABC in two equal parts. So, \frac{112}{2} = 56

3. m∠CBG=  56° because ∠ABG and ∠CBG are equal.

4. m∠DEG=  56° because ∠ABG ≅ ∠DEG


7 0
2 years ago
Read 2 more answers
A stadium brings in $16.25 million per year. it pays football-related expenses of $13.5 million and stadium expenses of $2.7 mil
Ganezh [65]
We have that

Profit margin is calculated by finding the net profit as a percentage of the revenue

Profit margin = [Net profit / Revenue]

Net profit= [Revenue-Cost]

we know that

Revenue = $16.25 million

Cost = $13.5 million + $2.7 million

Net profit = [16.25 million - (13.5 million + 2.7 million)]

Net profit = $0.05 million

Profit margin = 0.05 / 16.25

Profit margin = 0.003077 or 0.3077%


3 0
2 years ago
Two production lines are used to pack sugar into 5 kg bags. Line 1 produces twice as many bags as does line 2. One percent of ba
enyata [817]

Answer:

P(Bag is Defective) = 0.0167

Step-by-step explanation:

Line 1 produces twice as many bags as line 2. Let x be the number of bags produced by line 2.

No. of bags produced by line 2 = x

No. of bags produced by line 1 = 2x

Probability that the bag has been produced by line 1 can be written as:

P(Line 1) = No. of bags produced by line 1/Total no. of bags

             = 2x/(x+2x)

             = 2x/3x

P(Line 1) = 2/3. Similarly,

P(Line 2) = x/3x

P(Line 2) = 1/3

1% bags produced by line 1 are defective so the probability of line 1 producing a defective bag is:

P(Defective|Line 1) = 0.01

3% of bags from line 2 are defective, so:

P(Defective|Line 2) = 0.03

b. The probability that the chosen bag is defective can be calculated through the conditional probability formula:

P(A|B) = P(A∩B)/P(B)

<u>P(A∩B) = P(A|B)*P(B)</u>

The chosen defective bag can be either from line 1 or from line 2. So, the probability that the chosen bag is defective is:

P(Bag is Defective) = P(Defective and from Line 1) + P(Defective and from Line 2)

                                = P(D∩Line 1) + P(D∩Line 2)

                                = P(Defective|Line 1)*P(Line 1) + P(Defective|Line 2)*P(Line 2)

                                = (0.01)*(2/3) + (0.03)(1/3)

P(Bag is Defective) = 0.0167

7 0
2 years ago
Read 2 more answers
Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll
JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

So, the possibility of getting a total of 7 = \frac{6}{36}=\frac{1}{6}

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = \frac{2}{36} = \frac{1}{18}

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = \frac{28}{36} = \frac{7}{9}

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11

\frac{11}{3}+\frac{11}{3}-\frac{77}{9}

\frac{22}{3}-\frac{77}{9}

\frac{66-77}{9}

-\frac{11}{9}

-1.22

Therefore the expected payoff for this game is -$1.22.

4 0
2 years ago
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