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irina1246 [14]
3 years ago
6

bob loves his cows. He has 53 cows on his farm. Bob spends 10 minutes every day petting each cow. He spends 200 minutes twice a

week grooming the cows. He spends an extra 5 minutes every day to feed one cow. How much time does he spend taking care of his cows each week
Mathematics
1 answer:
ValentinkaMS [17]3 years ago
4 0
Bob has 53 cows and spends 10 min each day petting each cow.This is 530 minutes one day which is 3710 minutes in a week. 200 minutes twice a week is 400 minutes. 53 times extra 5 minutes is 265 everyday times 7 days a week which is 1855.

3710+400+1855=5965 minutes
99 hours and 25 minutes

Bob sure loves his cows...
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Answer:

0.288

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 \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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 17.48, \sigma = 3.25, n = 10, s = \frac{3.25}{\sqrt{10}} = 1.027740

Find the probability that the mean printing speed of the sample is greater than 18.06 ppm.

This is 1 subtracted by the pvalue of Z when X = 18.06. So

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

By the Central Limit Theorem

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

Z = \frac{18.06 - 17.48}{1.027740}

Z = 0.56

Z = 0.56 has a pvalue of 0.712

1 - 0.712 = 0.288

The answer is 0.288

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