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3 years ago

Each week as part of her workout, Monica walks 0.5 mile, runs 1.6 miles, and then walks 0.75 mile.

Mathematics

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

She will walk and run 148.2 miles in a year

Step-by-step explanation:

She walks: 0.75+0.5=1.25 miles each week

And: 1.25 *52=65 miles a year

She runs : 1.6 miles a week

And: 1.6*52 =83.2 miles a year

Total miles : 83.2+65=148.2 miles

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V=1/2pir^2h<br> Solve for h

Irina-Kira [14]

I think it would be h=V/(1/2pir^2)

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3 years ago

The mean cost of a five pound bag of shrimp is 50 dollars with a variance of 64. If a sample of 43 bags of shrimp is randomly se

anyanavicka [17]

Answer:

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

Step-by-step explanation:

To solve this question, we need 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(which is the square root of the variance) $\sigma$ , 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 50, \sigma = \sqrt{64} = 8, n = 43, s = \frac{8}{\sqrt{43}} = 1.22$

What is the probability that the sample mean would differ from the true mean by greater than 1 dollar?

Either it differs by 1 dollar or less, or it differs by more than one dollar. The sum of the probabilities of these events is decimal 1.

Probability it differs by 1 dollar or less:

pvalue of Z when X = 50+1 = 51 subtracted by the pvalue of Z when X = 50 - 1 = 49.

X = 51

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

By the Central Limit Theorem

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

$Z = \frac{51-50}{1.22}$

$Z = 0.82$

$Z = 0.82$ has a pvalue of 0.7939

X = 49

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

$Z = \frac{49-50}{1.22}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061

0.7939 - 0.2061 = 0.5878

Probability it differs by more than 1 dollar:

p + 0.5878 = 1

p = 0.4122

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

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4 years ago

(a) It takes the Earth 365 days to complete a full orbit around the sun. It takes Jupiter approximately 4,332 days, 14 h, and 9

kogti [31]

1 year = 365 days.

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First convert the hours and minutes in Jupiter orbit to days:

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849 / 60 = 14.15 hours.

14.15 / 24 = 0.59 days.

Jupiter's orbit is 4332.59 days.

Now divide that by 365 for the number of years:

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Answer:

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