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saul85 [17]
3 years ago
11

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. T

he critical value for a 98% confidence interval around this sample mean is ________.
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

153.158 ; 177.642

Step-by-step explanation:

Given the following:

Sample mean (m) = 165.40

Sample standard deviation (s) = 21.70

Sample size (n) = 17

α = 98%

Confidence interval = m ± z(SE)

z at 98% = 2.326

SE = s/√n

SE = 21.70/√17 = 5.2630230

Hence,

Confidence interval = 165.40 ± 2.326(5.2630230)

165.40 - 12.241791498 OR 165.40 + 12.241791498

153.158 ; 177.642

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Answer:

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4 years ago
Simplify rational expression
zubka84 [21]

Answer:

<u>The final answer is 35n/17r</u>

Step-by-step explanation:

Let's simplify the expression, this way:

1. Start with the coefficients:

(10 * 42)/(6 * 34)

We cancel 6 in the denominator and change 42 by 7. Now we have:

(10 * 7)/34

We now divide by 2 both in the numerator and the denominator. Now we have:

(5 * 7)/17.

The remaining numbers are prime numbers and it's not possible to continue simplifying, so we got:

35/17

2. Variable r:

We have r⁴/r⁵ = r ⁴⁻⁵ = r⁻¹ = 1/r¹ = 1/r

3. Variable n:

We have n⁴/n³ = n ⁴⁻³ = n¹/1 = n

4. Adding the pieces:

We put coefficients and variables together: 35/17 * 1/r * n

<u>The final answer is 35n/17r</u>

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4 years ago
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Murljashka [212]
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Digiron [165]

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