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Ahat [919]
3 years ago
5

There is 192 inches how much does 192 equals feet

Mathematics
2 answers:
alexira [117]3 years ago
7 0
1 ft = 12 in
192/12 = 16
16 ft = 192 in
sladkih [1.3K]3 years ago
6 0
Okay so, what you need to do is to first divide 192 by 12, since one foot equals 12 inches.

You should get 16 feet in total.
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A bus leaves Dhaka on Saturday night and is supposed to arrive at Cox’s Bazar at 08 17
guajiro [1.7K]

Answer:

9:34 pm

Step-by-step explanation:

First:

8:17 am

Go back 12 hours

8:17 pm

Round to 8:20

(keep the 3) Minus 43-17=26(3 is in this)

Therefore, 9:34 pm

Sorry if this doesn't make sense, it's hard to explain haha

7 0
2 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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Answer:

59

Step-by-step explanation:

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Step-by-step explanation:

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This only happens when the ordinary ratio is greater than one, the terms of the sequence will get bigger and bigger, and if you add larger numbers you won't get a definitive answer.
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