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Aleks [24]
3 years ago
11

Stephon makes the following statements: statement 1 lim x -3 f(x) exists and is equal to 1 statement 2 lim x 1 f(x) exists and i

s equal to 1

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

Neither statement 1, nor statement 2 are correct

The given Stephon's statements are;

Statement 1; \mathbf{\lim \limits  _{x \to -3} f(x)}  \ \mathbf{Exist} \ and \mathbf{\lim \limits  _{x \to -3} f(x) = 1}

Statement 2; \mathbf{\lim \limits  _{x \to 1} f(x)}  \ \mathbf{Exist} \ and \mathbf{\lim \limits  _{x \to 1} f(x) = 1}

The analysis of the graph and reason for the answer

From the graphed line on the left of the y-axis, we have an open circle at x = -3, and an arrow at the other end pointing towards negative infinity, (-∞) which indicates that the domain is -∞ ≤ x < -3, therefore, at x = -3, f(x) does not exist, therefore, we can write the following statement

The limits of the domain and range of the graph includes;

\mathbf{\lim \limits  _{x \to -3} f(x)} = \mathbf{Does \ not \ exist}

f(x) = Defined for -∞ ≤ x < -3

Similarly, from the graphed line on the right of the y-axis, we have an open circle at x = 1 and an arrow at the other end of the line f(x) = 4 pointing towards positive (+∞) infinity, which indicates the domain and the graph of the function is 1 < x ≤ ∞ , therefore, f(x) does not exist at x = 1, and we can write

\mathbf{\lim \limits  _{x \to 1} f(x)} = \mathbf{Does \ not \ exist}

From we above, we have that neither statement 1, nor statement 2 are correct

Learn more about open and closed circles on graph lines

brainly.com/question/8648835

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Emma has $0.95 in dimes and nickles. She has five more dimes than nickles. How many coins of each type does she have?
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Answer:

3 nickels and 8 dimes

Step-by-step explanation:

3 nickels equals 15

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so ur answer is 3 nickels and 8 dimes

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3 years ago
5) Two machines M1, M2 are used to manufacture resistors with a design
Basile [38]

Answer:

Since M1 has the higher probability of being in the desired range, we choose M1.

Step-by-step explanation:

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.

Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.

So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.

For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.

M1:

Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that \mu = 1050, \sigma = 100

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

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

Z = \frac{1100 - 1050}{100}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 900

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

Z = \frac{900 - 1050}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.6915 - 0.0668 = 0.6247.

M1 has a 62.47% probability of being in the desired range.

M2:

M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that \mu = 1000, \sigma = 120

X = 1100

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

Z = \frac{1100 - 1000}{120}

Z = 0.83

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X = 900

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

Z = \frac{900 - 1000}{120}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

0.7967 - 0.2033 = 0.5934

M2 has a 59.34% probability of being in the desired range.

Since M1 has the higher probability of being in the desired range, we choose M1.

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