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Oliga [24]
3 years ago
11

A quantity with an initial value of 870 decays exponentially at a rate of 0.55% every 4 decades. What is the value of the quanti

ty after 85 decades, to the nearest hundredth?

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Step-by-step explanation:

Below is an attachment containing the solution.

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What are the equations
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Answer:

(-3,4)

(-1,4)

(-3,7)

Step-by-step explanation:

(1,1) → (1-4, 1+3) → (-3,4)

(3,1) → (3-4, 1+3) → (-1,4)

(1,4) → (1-4, 4+3) → (-3,7)

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The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel
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Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation \sigma = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

S.E = \frac{\sigma}{\sqrt{n}}

= \frac{2}{\sqrt{45}}

= \frac{2}{6.7082}

SE = 0.298

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

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