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3 years ago

How many 8's are in the answer and sum 5890/45?

Mathematics

1 answer:

Novay_Z [31]3 years ago

3 0

Answer:

1. There are infinite 8's

2. The estimate is 130.89

3. The difference is 0.01

Explanation:

There are countless, or infinite, 8's in the answer because the result is a periodic decimal:

$\dfrac{5890}{45}=130.8888 . . .$

The notation used for the periodic decimal numbers is an arc over the number that is repeated. In this case, it would be an arc over the first 8.

I will use a bar just to show how it is, but instead of a bar it should be an arc:

$\dfrac{5890}{45}=130.\overline 8$

The estimation or approximation to two decimal places requires to drop the 8's since the third one and increase the second 8 to 9 (because the first decimal dropped is greater than or equal to 5).

Thus the estimated value would be 13.89.

The difference is:

130.89 - 130.888. . . = 0.0111. . .
Rounded to 2 decimal places: 0.01

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Suppose a sample of 100 families of four vacationing at Niagara Falls resulted in sample mean of $282.45 spent per day and a sam

Juli2301 [7.4K]

Given Information:

Mean = μ = $282.45

Standard deviation = σ = $64.50

Sample size = n = 100

Confidence level = 95%

Required Information:

95% Confidence interval = ?

Answer:

95% Confidence interval = ($269.81, $295.09)

Step-by-step explanation:

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

The confidence interval is given by

$CI = \bar{x} \pm MoE\\$

Where $\bar{x}$ is the mean and MoE is the margin of error given by

$MoE = z_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\$

Where σ is the standard deviation, n is the sample size and $z_{\alpha/2}$ is the z-score corresponding to 95% confidence level.

$z_{\alpha/2} = 1 - 0.95 = 0.05/2 = 0.025\\\\z_{0.025} = 1.96$

$MoE = 1.96\cdot \frac{64.50}{\sqrt{100} } \\\\MoE = 1.96\cdot 6.45\\\\MoE = 12.64\\$

Finally, the confidence interval is

$CI = \bar{x} \pm MoE\\\\CI = 282.45 \pm 12.64\\\\CI = 282.45 - 12.64 \: and \: 282.45 + 12.64\\\\CI = \$269.81 \: and \:\:\$295.09\\$

Therefore, we are 95% sure that the true population mean amount spent per day by a family of four visiting Niagara Falls is within the interval of ($269.81, $295.09)

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