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vazorg [7]
3 years ago
13

Peanuts cost £6.40 per kg. Find the cost of 100 g of peanuts

Mathematics
2 answers:
I am Lyosha [343]3 years ago
8 0

Answer:

1 \: kg \: cost \: 6.40 \\ so \: 100kg \: cost \: 6.40 \times 100 = 640 \\1000g \:  =  > 1kg \\ 100g =  > .1kg \\ then \: for \: .1g \: 640 \times  \div 100 \times .1 = .640 \\ hank \: you

zaharov [31]3 years ago
4 0

Step-by-step explanation:

Solution,

  • Cost of 1 gram of peanuts=£6.40
  • Cost of 100 grams of peanuts =

£6.40×100

£640

Hence, the cost of 100 grams of peanuts is £640.

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Suppose that in one region of the country the mean amount of credit card debt perhousehold in households having credit card debt
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Answer:

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 15250, \sigma = 7125, n = 1600, s = \frac{7125}{\sqrt{1600}} = 178.125

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly

This probability is the pvalue of Z when X = 1600 + 300 = 1900 subtracted by the pvalue of Z when X = 1600 - 300 = 1300. So

X = 1900

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

By the Central Limit Theorem

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

Z = \frac{1900 - 1600}{178.125}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535.

X = 1300

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

Z = \frac{1300 - 1600}{178.125}

Z = -1.68

Z = -1.68 has a pvalue of 0.0465.

0.9535 - 0.0465 = 0.907.

The probability that the mean amount of credit card debt in a sample of 1600 such households will be within $300 of the population mean is roughly 0.907 = 90.7%.

7 0
4 years ago
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