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andre [41]
3 years ago
5

If n erasers have a weight of 80 grams, what is the total weight of 50 erasers?

Mathematics
2 answers:
Black_prince [1.1K]3 years ago
6 0

Answer:

The\ weight\ of\ the\ 50\ erasers\ be\ \frac{4000}{n}\ grams.

Step-by-step explanation:

As given

if n erasers have a weight of 80 grams .

I.e

n erasera =  80 grams.

1 erasers weight .

1\ erasers\ weight = \frac{80}{n}

Now find out the weight of the 50 erasers.

Weight\ of\ 50\ erasers = \frac{50\times 80}{n}\ grams

Simplify the above

Weight\ of\ 50\ erasers = \frac{4000}{n}\ grams

Therefore\ the\ weight\ of\ the\ 50\ erasers\ be\ \frac{4000}{n}\ grams.

Sergio [31]3 years ago
3 0

Answer:

weight of 50 erasers is \frac{80}{n}\times 50  g

Step-by-step explanation:

Given that n erasers have a weight of 80 grams,

we have to find total weight of 50 erasers

Now, weight of n erasers=80 g

∴ weight of 1 eraser = \frac{80}{n} g

⇒  weight of 50 erasers =  \frac{80}{n}\times 50  g

Hence, weight of 50 erasers is  \frac{80}{n}\times 50  g

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A statistician calculates that 8% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the
hichkok12 [17]

Answer:

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

A statistician calculates that 8% of Americans own a Rolls Royce.

This means that p = 0.08

Sample of 595:

This means that n = 595

Mean and standard deviation:

\mu = p = 0.08

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*0.92}{595}} = 0.0111

What is the probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%?

Proportion above 8% + 3% = 11% or below 8% - 3% = 5%. Since the normal distribution is symmetric, these probabilities are equal, and so we find one of them and multiply by 2.

Probability the proportion is less than 5%:

P-value of Z when X = 0.05. So

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

By the Central Limit Theorem

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

Z = \frac{0.05 - 0.08}{0.0111}

Z = -2.7

Z = -2.7 has a p-value of 0.0035

2*0.0035 = 0.0070

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

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It's 3h+3 hope this helps
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