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Elodia [21]
3 years ago
13

A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviatio

n of 0.79. calculate a 95% large-sample ci for the true average percentage elongation μ. (give answers accurate to 2 decimal places.)
Mathematics
1 answer:
ss7ja [257]3 years ago
5 0

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

 

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

 

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

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How do I solve this
Veronika [31]

Answer:

P(A or B) = 1.16

Step-by-step explanation:

Given probability:

Probability of event A = P(A) = 0.46

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P(A or B)

Computation:

If A is an incident and B is a separate event, P(A or B) is the possibility of either A, B, or both events occurring.

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Sergio [31]

Radius = 5 cm

Height = 8 cm

Volume = πr²h

Volume = π(5 cm)² × 8 cm

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6 0
3 years ago
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PLEASE HELP!! will give brainliest
frutty [35]
A:
Table 1:
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X= cups so
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Note: another way instead of f(x)=kx is y=kx
B:
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T1: 8
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So, looking at our slopes, table 3 has the greatest slope because it has the greater slope out of the other two.
C:
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Answer:

Step-by-step explanation:

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3 years ago
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