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jolli1
1 day ago
10

Anya found the slope of the line that passes through the points (–7, 4) and (2, –3). Her work is shown below. Let (x2, y2) be (–

7, 4) and (x1, y1) be (2, –3). m = = = The slope is . What error did she make? She simplified the denominator incorrectly. The denominator simplifies to –7. She labeled the points incorrectly. The point (–7, 4) should be (x1, y1). She used an incorrect formula. The formula should be the change in y-values with respect to the change in the x-values. She used an incorrect formula. The formula should be the sum of the x-values with respect to the sum of the y-values.
Mathematics
2 answers:
AnnZ [9.1K]1 day ago
8 0
C. Anya employed an incorrect formula. The correct formula must reflect the change in the y-values in relation to the change in the x-values.
zzz [9K]1 day ago
6 0

Answer:

The accurate formula should involve the change in y-values correlating to the change in the x-values.

Step-by-step explanation:

It is given that the slope of the line through two points was calculated based on the provided work.

We must identify the mistake in the method

The slope of the line between two points = \frac{y_2-y_1}{x_2-x_1}

Let (x2, y2) represent (–7, 4) and (x1, y1) represent (2, –3).

Thus, the numerator equals 4 - (-3) = 7

and the denominator equals -7 - 2 = -9

The formula should indeed be the change in y-values correlating to the change in x-values.

Utilizing just this formula will yield the correct slope.

However, Anya had misapplied the formula.

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1 month ago
Let X denote the data transfer time (ms) in a grid computing system (the time required for data transfer) between a "worker" com
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Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

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The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

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=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176

6 0
12 days ago
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9 days ago
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A = x (x-4) -0.5 (8) (x - 4) - 4 (8)

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20 days ago
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