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rjkz [21]
2 years ago
13

Write each repeating decimal using bar notation. 2.034034 0.9222 0.7777

Mathematics
2 answers:
DochEvi [55]2 years ago
6 0
2,0,3,4,7 are the ones repeating
the most repeating ones are 2,0, 7
zubka84 [21]2 years ago
5 0

Answer:

4,2,7

Step-by-step explanation

they are the ones repeating at the end

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A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard
Sauron [17]

Answer:

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = $393.50

Standard deviation r = $50.30

Number of samples n = 25

Confidence interval = 95%

z value(at 95% confidence) = 1.96

Substituting the values we have;

$393.50+/-1.96($50.30/√25)

$393.50+/-1.96($10.06)

$393.50+/-$19.7176

$393.50+/-$19.72

= ( $373.78, $413.22)

Therefore, the 95% confidence interval (a,b) = ($373.78, $413.22)

5 0
2 years ago
The following observations are on stopping distance (ft) of a particular truck at 20 mph under specified experi- mental conditio
zzz [600]

Answer:

see explaination

Step-by-step explanation:

Data : 32.1 , 30.6 , 31.4 , 30.4 , 31.0 , 31.9

Mean X-bar = 31.23

SD = 0.689

a)

Null Hypothesis : Xbar = mu

Alternate Hypothesis : Xbar > mu

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 30 ) /(0.689/sqrt(6))

= 4.372

P-value = ~0

Since P-vale < 0.01 , we will reject null hypothesis.

The data suggest that true average stopping ditance exceeds the maximum.

b)

i) SD = 0.65

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.65/sqrt(6))

= 0.867

P-value = 0.3859 Answer

ii) SD = 0.65

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.65/sqrt(6))

= -2.9

P-value = 0.0037 Answer

c)

i) SD = 0.8

mu = 31

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 31) /(0.8/sqrt(6))

= 0.704

P-value = 0.4814 Answer

ii) SD = 0.8

mu = 32

z = (Xbar - mu) / (SD/sqrt(n))

= (31.23 - 32) /(0.8/sqrt(6))

= -2.357

P-value = 0.0184 Answer

The probabilities obtained in part c are comparatively higher than that of part b.

d)

i) For alpha =0.01

z = (Xbar - mu) / (SD/sqrt(n))

=> -2.32 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-2.32

=> sqrt(n) = (0.65*(-2.32)) / (31.23 - 31)

=> n = 43 Answer

ii) For beta =0.10

z = (Xbar - mu) / (SD/sqrt(n))

=> -1.28 = (31.23 - 31) /(0.65/sqrt(n))

=> (0.65/sqrt(n)) = (31.23 - 31)/-1.28

=> sqrt(n) = (0.65*(-1.28)) / (31.23 - 31)

=> n = 13 Answer

4 0
2 years ago
which of the sets shown includes the elements of set Z that both negative numbers and multiples of ?. Z={-34,-28,-16,-2,4,8,12,2
Ronch [10]

Answer:

multiples of 2

Step-by-step explanation:

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2 years ago
Match the following STATEMENTS to the reasons listed. NOTE: In a traditional proof format, the statements would be on the left s
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1. RM = SN, TM = TN Addition Property of Equality  

2. ∠T = ∠T Reflexive  

3. RM + TM = SN + TN Substitution  

4. RM + TM = RT, SN + TN = ST Betweeness  

5. RT = ST CPCTE  

6. Triangle RTN congruent to Triangle STM Given  

7. RN = SM SAS

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2 years ago
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The table shows different quantities of flour and the number of cookies that can be made with each quantity of flour. If the rel
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2 years ago
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