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Finger [1]
3 years ago
12

A study found that 9% of dog owners brush their dog’s teeth. Of 578 dog owners, about how many would be expected to brush their

dog’s teeth? Explain. A. 50; 9% ≈ 10% and 578 should be rounded down to 500; 10% of 500 = 50 B. 60; 9% ≈ 10% and 578 ≈ 600; 10% of 600 = 60 C. 70; 9% ≈ 10% and 578 ≈ 700; 10% of 700 = 70 D. 100; 9% ≈ 10% and 578 should be rounded up to 1,000; 10% of 1,000 = 100
Mathematics
1 answer:
Amanda [17]3 years ago
5 0

Answer:

A. 50

Step-by-step explanation:

Simply multiply 9% and 578 to get your answer:

578(0.09) = 52.02

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Step-by-step explanation:

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A student desired to invest $1,540 into an investment at 9% compounded semiannually for 6 years. With all else equal, what is th
irga5000 [103]

Answer:

The future value of this initial investment after the six year period is $2611.6552

Step-by-step explanation:

Consider the provided information.

A student desired to invest $1,540 into an investment at 9% compounded semiannually for 6 years.

Future value of an investment: FV=P(1+r)^n

Where Fv is the future value, p is the present value, r is the rate and n is the number of compounding periods.

9% compounded semiannually for 6 years.

Therefore, the value of r is: r=\frac{0.09}{2}=0.045

Number of periods are: 2 × 6 = 12

Now substitute the respective values in the above formula.

FV=1540(1+0.045)^{12}

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FV=1540(1.69588)

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The mean per capita income is 16,127 dollars per annum with a variance of 682,276. What is the probability that the sample mean
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Answer:

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

The standard deviation is the square root of the variance. So

\mu = 16127, \sigma = \sqrt{682276} = 826, n = 476, s = \frac{826}{\sqrt{476}} = 37.86

What is the probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

Either it differs by 104 or less dollars, or it differs by more than 104 dollars. The sum of the probabilities of these events is 100. I am going to find the probability that it differs by 104 or less dollars first.

Probability that it differs by 104 or less dollars first.

pvalue of Z when X = 16127 + 104 = 16231 subtracted by the pvalue of Z when X = 16127 - 104 = 16023. So

X = 16231

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{16231 - 16127}{37.86}

Z = 2.75

Z = 2.75 has a pvalue of 0.9970

X = 16023

Z = \frac{X - \mu}{s}

Z = \frac{16023 - 16127}{37.86}

Z = -2.75

Z = -2.75 has a pvalue of 0.0030

0.9970 - 0.0030 = 0.9940

99.40% probability that it differs by 104 or less.

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p + 99.40 = 100

p = 0.60

0.60% probability that the sample mean would differ from the true mean by more than 104 dollars if a sample of 476 persons is randomly selected

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