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3 years ago

What is the answer to 5t-26=18t

Mathematics

2 answers:

AlladinOne [14]3 years ago

6 0

Answer:

t = -2

Step-by-step explanation:

5t - 26 = 18t

Add 26 to both sides:

5t - 26 + 26 = 18t + 26

Simplify:

5t = 18t + 26

Subtract 18t from both sides:

5t - 18t = 18t + 26 - 18t

Simplify:

-13t = 26

Divide both sides by -13:

-13t/-13 = 26/-13

Therefore:

t = -2

I hope this helps!

boyakko [2]3 years ago

6 0

You want to isolate t as much as possible

Subtract 18t from not sides and add 26 to both sides

-13t=26 | divide each side by -13
t=-2

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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

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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.

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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 $X = 204.1$ .

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

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

$Z = 3.14$

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