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Nady [450]
2 years ago
9

HELP ME PLEEEEEEAAAASSSSEEEEE

Mathematics
1 answer:
Anettt [7]2 years ago
5 0

Answer:

Step-by-step explanation:

Remark

There are a couple of things you should keep in mind.

1. Your coefficient should always be a number between 1 and 9.999...

2. If you move right, you have a negative exponent. If you move left, you get a positive exponent.

Solution

So you are going right. The exponent is minus. You have 2 places to go to get a number between 1 and 9.99999. In this case, the coefficient is 3.55 So it goes in the blue box.

The red box is - 2

The answer is: blue 3.55

Orange: -2

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future \: value \: minus \: interest

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Add. 34.700 + 98.428 98.775 132.435 132.980 133.128
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34.700+98.42+98.775+132.435+132.980+133.128

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6 eggs weigh 3/4 of a pound. How much does each egg weigh?
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3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7

This probability is 1 subtracted by the pvalue of Z when X = 204.1.

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

Z = \frac{204.1 - 201.9}{0.7}

Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

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