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mr Goodwill [35]
4 years ago
14

Sue buys 5 pieces of fabric.Each piece is 1 7/10 yards long.What is the total length of the fabric she buys?One yard of the fabr

ic costs $5.How much does she pay for all 5 pieces of fabric?
Mathematics
1 answer:
irina [24]4 years ago
5 0

Answer:

The total length of fabric sue buys is 8.5 yards and total cost of the fabric is $42.5 .

Step-by-step explanation:

As given

Sue buys 5 pieces of fabric.

Each\ piece\ is\ 1 \frac{7}{10}\ yards\ long.

i.e

Each\ piece\ is\ \frac{17}{10}\ yards\ long.

Total\ length\ of\ fabric\ sue\ buys = \frac{5\times 17}{10}\ yards

Total\ length\ of\ fabric\ sue\ buys = \frac{85}{10}\ yards

In the decimal form

Total length of fabric sue buys = 8.5 yards

As given

One yard of the fabric costs $5

Total\ cost\ of\ fabric\ sue\ buys = 8.5\times 5

                                                            = $ 42.5  

Therefore the total length of fabric sue buys is 8.5 yards and total cost of the fabric is $42.5 .


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Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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 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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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Solve the following system of linear equation by graphing. <br>4x + 2y = -2 <br>2x – 3y = 9 ​
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