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Margarita [4]
2 years ago
9

FIRST PERSON TO ANSWER GETS BRAINLIEST OR FIVE STARS Define the domain and range of the function.

Mathematics
2 answers:
uysha [10]2 years ago
7 0

Answer:

Domain: (-∞, ∞)

Range: (-∞, ∞)

Step-by-step explanation:

The domain are the x-values included in the function (the horizontal axis).

The range are the y-values included in the function (the vertical axis).

The two arrows on the ends of the line (pointing upwards and downwards respectively) indicate that the function goes in those direction for infinity. Therefore, if there are an infinite amount of y-values, the range is (-∞, ∞).

While the slope is quite steep, there is still a slope and slowly "expands" the line on the horizontal axis. Because there is no limit to the y-values, the domain will also expand infinitely. Therefore, the domain is also (-∞, ∞).

igomit [66]2 years ago
3 0

Answer:

Domain: All real numbers (interval notation: (-∞, ∞)

Range: All real numbers (interval notation: (-∞, ∞).

Step-by-step explanation: The domain of a graph is all the x-values where the line touches or will eventually touch.

The range of a graph is like the domain, but it is all the y-values where the line touches or will touch.

In a linear function in the form y = mx + b or ax + by = c, the domain and range will always be "all real numbers" and the interval will always be (-∞,  ∞). This makes sense since a line always goes on forever (unless it's piecewise or an inequality.) So the line will eventually meet every single x and y value on the xy graph.

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Find the equation of a line which passes through (2 4) with a slope of ½​
Norma-Jean [14]

Answer:

The equation of the straight line is  x - 2y +6 =0

Step-by-step explanation:

<u>Explanation:</u>-

Given a point ( 2, 4) and slope m =  \frac{1}{2}

The equation of the straight line passing through the point and having slope 'm'

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Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your ans
Alenkinab [10]

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

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 normally distributed samples are 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.

In this problem, we have that:

\mu = 50, \sigma = 1.8

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?

Here n = 17, s = \frac{1.8}{\sqrt{17}} = 0.4366

This probability is 1 subtracted by the pvalue of Z when X = 51. So

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

By the Central Limit Theorem

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

Z = \frac{51 - 50}{0.4366}

Z = 2.29

Z = 2.29 has a pvalue of 0.9890

1 - 0.989 = 0.011

0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?

Here n = 17, s = \frac{1.8}{\sqrt{45}} = 0.2683

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

Z = \frac{51 - 50}{0.0.2683}

Z = 3.73

Z = 3.73 has a pvalue of 0.9999

1 - 0.9999 = 0.0001

0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

8 0
4 years ago
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