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aleksandr82 [10.1K]
2 years ago
13

The weight of each fountain pen manufactured in a factory must be 10g with a tolerance of 2g. Pens that are not within the toler

ated weight must be thrown away. Write an inequality that can be used to assess which pens are tolerable (w is the weight of the pen)?
Mathematics
1 answer:
FromTheMoon [43]2 years ago
5 0

According to the information given, the absolute value inequality that can be used to assess which pens are tolerable is:

|w - 10| \leq 2

The <em>absolute value</em> function measures the <u>distance of a point to the origin</u>, and is defined by:

  • |x| = x, x \geq 0
  • |x| = -x, x < 0

The weight must be of 10g, with a tolerance of 2g, which means that the <u>absolute value of the difference between the weight w and 10g must be of at most 2g</u>, hence the inequality is:

|w - 10| \leq 2

To learn more about absolute value inequalities, you can take a look at brainly.com/question/24005819

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Since it is on the left up corner, its in quadrant 2
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Identify the range of the function shown in the graph.
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The range of the function shown in the graph are all real numbers.

<h3>What is the range of a function?</h3>

The range of a function is the set of its possible output values.

The range of a function are the dependent variables.

In other words, the range of a function is the complete set of all possible resulting values of the dependent variable (y), after we have substituted the domain or independent variable.

According to the values of y on the graph the range are all real numbers.

learn more on range here: brainly.com/question/12101011

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8 0
2 years ago
Which equation represents a line that is parallel to y=-4x+3 and that passes through the point (-1, 2)?
belka [17]
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Put x=-1 in each of C and D and we get y=2 for C and y=1 for D. Since the point is (-1,2) the answer is C.
6 0
4 years ago
A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at t
bogdanovich [222]

Answer:

The critical value is T = 4.604.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Step-by-step explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

14,11,18,9,19

Sample mean is sum of all values divided by the number of values. Thus:

\overline{x} = \frac{14+11+18+9+19}{5} = 14.2

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

s = \sqrt{\frac{(14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2}{5}} = 3.868

Confidence interval:

We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 5 - 1 = 4

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 4.604.

The critical value is T = 4.604.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 4.604\frac{3.868}{\sqrt{5}} = 7.964

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

8 0
3 years ago
A hat is originally priced at $22 and now it is $18. What percent is the sales price of the total bill?
Bogdan [553]

So this means that you are asking $18 is what percent of $22.

First: Divide 18 by 22.

Second: You get the repeating decimal of 0.81818181.

Third: Find the first 2 decimals and use it, so the answer is approximately 81%.

7 0
3 years ago
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