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Lisa [10]
3 years ago
6

(Please someone help me!) (No links!)

Mathematics
1 answer:
Nataly_w [17]3 years ago
8 0

Answer:

3 cups

Step-by-step explanation:

36 : 24 = 4.5 : x

where x is the amount of flour he should use

Therefore,

\frac{36}{24} = \frac{4.5}{x}

Cross multiply:  36x = 4.5 \times 24

\implies 36x = 108

Divide both sides by 36:  x = 108 \div 36 = 3

Therefore, he should use 3 cups of flour

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The plot that organizes the data into 4 groups of equal sizes is box and whisker plot.

The image below shows a box and whisker plot. Following are the elements of box and whisker plot:

Minimum = This is the smallest value of the data set

Q1 = First (Lower) Quartile of the data set. 25% of the data values lie below this point

Q2 = Second Quartile or Median. This is the central value so 50% of the data values lie below this point

Q3 = Third (Upper) Quartile of the data set. 75% of the data values lie below this point.

Maximum = This is the maximum value of the data set.

Based on box and whisker plot we can compare two or more sets of data by comparing the spread of the data. We can also directly observe from the box and whisker plot if the data is uniform, normal or skewed. Using box and whisker plot we can also visualize any outliers that may be in the data.

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What is the significance of a correlation coefficient of 0.88?
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What is the length of AC?<br> ОЗft<br> o4ft<br> 9 ft<br> 09 ft<br> o18 ft
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Step-by-step explanation:

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The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
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Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

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Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

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t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

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Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

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b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

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6700± 1.645 (500/√50)

6700±116.336

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