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Brrunno [24]
3 years ago
7

In order for the limit of F(X) to exist at x=a, which of the following must hold true

Mathematics
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

A.

Step-by-step explanation:

Given: limit of f(x) exists at x = a

To choose: the correct option

Solution:

A limit is a value that a function approaches as the input approaches some value.

\lim_{x\rightarrow a}f(x)=L means that as x approaches to value a then f(x) approaches to value L.

A limit of a function f(x) exists at x = a if \lim_{x\rightarrow a^-}f(x)=\lim_{x\rightarrow a^+}f(x)=L

So, option A. is correct.

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Step-by-step explanation:

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2 years ago
Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal(\mu=5.0,\sigma^{2} =1.7^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean thickness = 5.0 mm

           \sigma = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

    P(X < 3.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{3.0-5.0}{1.7} ) = P(Z < -1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

    P(X > 7.0 mm) = P( \frac{X-\mu}{\sigma} > \frac{7.0-5.0}{1.7} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X \leq 3.0 mm)

    P(X < 7.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{7.0-5.0}{1.7} ) = P(Z < 1.18) = 0.881

    P(X \leq 3.0 mm) = P( \frac{X-\mu}{\sigma} \leq \frac{3.0-5.0}{1.7} ) = P(Z \leq -1.18) = 1 - P(Z < 1.18)

                                                           = 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

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