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allochka39001 [22]
2 years ago
10

A shipment of ball bearings with a mean diameter of 25 mm and a standard deviation of 0.2 mm is normally distributed. By how man

y standard deviations does a ball bearing with a diameter of 25.6 mm differ from the mean? 0.6 1 2 3
Mathematics
1 answer:
Svetach [21]2 years ago
6 0
You need to calculate how many times the required difference is of the standard deviation, i.e. the ratio difference / standard deviation.


These are the calculations:


Standard deviation = 0.2 mm


Difference between 25.6mm and the mean = 25.6mm - 25mm = 0.6 mm


Ratio difference / standard deviation = 0.6mm / 0.2 mm = 3.


Then, the answer is that a ball with a diameter of 25.6 mm differs 3 standard deviations from the mean.
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Answer:

Sum = -81

Step-by-step explanation:

See the comment for complete question.

Given

c = 36 ----- Constant

No coefficient of x^2

Required:

Determine the sum of all distinct positive integers of the coefficient of x

Reading through the complete question, we can see that the question has 3 terms which are:

x^2 ---- with no coefficient

x ---- with an unknown coefficient

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So, the equation can be represented as:

x^2 + ax + 36 = 0

Where a is the unknown coefficient

From the question, we understand that the equation has two negative integer solution. This can be represented as:

x = -\alpha and x = -\beta

Using the above roots, the equation can be represented as:

(x + \alpha)(x + \beta) = 0

Open brackets

x^2 + (\alpha + \beta)x + \alpha \beta = 0

To compare the above equation to x^2 + ax + 36 = 0, we have:

a = \alpha + \beta

\alpha \beta = 36

Where: \alpha, \beta and \alpha \ne \beta

The values of \alpha and \beta that satisfy \alpha \beta = 36 are:

\alpha = -1 and \beta = -36

\alpha = -2 and \beta = -18

\alpha = -3 and \beta = -12

\alpha = -4 and \beta = -9

So, the possible values of a are:

a = \alpha + \beta

When \alpha = -1 and \beta = -36

a = -1 - 36 = -37

When \alpha = -2 and \beta = -18

a = -2 - 18 = -20

When \alpha = -3 and \beta = -12

a = -3 - 12 = -15

When \alpha = -4 and \beta = -9

a = -4 - 9 = -13

At this point, we have established that the possible values of a are: -37, -20, -15 and -9.

The required sum is:

Sum = -37 -20 -15 - 9

Sum = -81

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