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3 years ago

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Ling worked three more hours on Tuesday than she did on Monday. On Wednesday, she worked one hour more than twice the number of

hours that she worked on Monday. The total number of hours is two more than five times the number of hours worked on Monday. Write and solve an equation to find the number of hours she worked on Monday.

2 answers:

Nina [5.8K]3 years ago

7 0

Let x be the number of hours ling work on monday.

We know that she worked three more hours on tuesday that in monday, this can be express as :

$x+3$

We also know that in wednesday she worked on more hour than twice the number on mondays, this can be expressed as:

$2x+1$

The total number of hours she worked this three days in two more than five the number of hours she worked on monday, this can be express as :

$5x+2\\$

Now , once we have all the expressions we add the expressions of the days and equate them to the total

$x+(x+3)+(2x+1)=5x+2$

Now we solve the equation

$x+(x+3)+(2x+1)=5x+2\\x+x+3+2x+1=5x+2\\4x+4=5x+2\\5x-4x=4-2\\x=2$

Therefore , she worked 2 hours on monday.

PLEASE MARK ME AS BRAINLIEST

serg [7]3 years ago

6 0

Answer:

2 hours (no work included)

Step-by-step explanation:

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Answer:

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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

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

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so $\mu = 0.5, \sigma = 0.05$ .

What is the probability that a line width is greater than 0.62 micrometer?

That is $P(X > 0.62)$

So

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

$Z = \frac{0.62 - 0.5}{0.05}$

$Z = 2.4$

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

$P(X \leq 0.62) + P(X > 0.62) = 1$

$P(X > 0.62) = 1 - 0.99180 = 0.0082$

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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Step-by-step explanation:

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faltersainse [42]

Answer:

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A is the correct option.

Step-by-step explanation:

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A is the correct option.

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