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n200080 [17]
3 years ago
10

Find the limit of the function by using direct substitution. (2 points) limit as x approaches two of quantity x squared plus eig

ht x minus two. 18 Does not exist 0 -18
Mathematics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

18

Step-by-step explanation:

Given f(x) = x^2 +8x - 2

applying limit as x approaches two

\lim_{x \to \2} _2 (x^2+8x-2)

substitute direct method put x =2

\lim_{x \to \2} _2 (x^2+8x-2) = 2^{2} +8(2)-2 =18

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Which expression is equivalent to 15 + 35
olya-2409 [2.1K]

Answer:

It says “which” so is it multiple choice??

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3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
kkurt [141]

Answer:

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean diameter of 144 millimeters, and a variance of 49.

This means that \mu = 144, \sigma = \sqrt{49} = 7

Sample of 46:

This means that n = 46, s = \frac{7}{\sqrt{46}}

Wat is the probability that the sample mean would differ from the population mean by more than 2 millimeters?

Above 144 + 2 = 146 or below 144 - 2 = 142. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them and multiply by two.

Probability the sample mean is below 142:

p-value of Z when X = 142, so:

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

By the Central Limit Theorem

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

Z = \frac{142 - 144}{\frac{7}{\sqrt{46}}}

Z = -1.94

Z = -1.94 has a p-value of 0.0262

2*0.0262 = 0.0524

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

8 0
3 years ago
Logan has a budget of $400 to have family pictures taken. There is a sitting fee of $38. Prints cost $25 per page. How many page
fredd [130]

Answer:

14

Step-by-step explanation:

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362/25=14.48

Logan can only print 14 whole pages.

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