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sasho [114]
3 years ago
13

Please answer this!?

Mathematics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

d = 50 \times 2 \\= 100

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HELPP please ??? I’ll appreciate it a lot .
victus00 [196]

Answer:   \sqrt{41}

On a keyboard you would type sqrt(41)

===========================================

Work Shown:

\text{point L} = (x_1, y_1) = (2,-1)\\\\\text{point N} = (x_2, y_2) = (7,-5)\\\\

d = \text{distance from L to N}\\\\d = \sqrt{ (x_1-x_2)^2 + (y_1-y_2)^2 } \ \text{ ... distance formula}\\\\d = \sqrt{ (2-7)^2 + (-1-(-5))^2 }\\\\d = \sqrt{ (2-7)^2 + (-1+5)^2 }\\\\d = \sqrt{ (-5)^2 + (4)^2 }\\\\d = \sqrt{ 25 + 16 }\\\\d = \sqrt{ 41 }\\\\

----------

A slight alternative is to plot the point M(2,-5) to form right triangle LMN. The 90 degree angle is at point M.

The legs are of length LM = 4 and MN = 5, which are found by subtracting the x coordinates together and the y coordinates together (or you can count the spaces). From there, use the Pythagorean theorem to get the hypotenuse LN.

The distance formula is an altered version of the Pythagorean theorem.

5 0
3 years ago
Read 2 more answers
The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an
Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
3 years ago
A local bank surveyed the status of ASU student accounts and found that the average overdraft was $21.22 with a standard deviati
svlad2 [7]

Answer: the probability of a student being overdrawn by more than $18.75 is 0.674

Step-by-step explanation:

Since the bank overdrafts of ASU student accounts are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = bank overdraft of Asu students.

µ = mean

σ = standard deviation

From the information given,

µ = $21.22

σ = $5.49

We want to find the probability of a student being overdrawn by more than $18.75. It is expressed as

P(x > 18.75) = 1 - P(x ≤ 18.75)

For x = 18.75,

z = (18.75 - 21.22)/5.49 = - 0.45

Looking at the normal distribution table, the probability corresponding to the z score is 0.326

Therefore,

P(x > 18.75) = 1 - 0.326 = 0.674

4 0
3 years ago
Please can you send on this question​
ella [17]

Answer:

C

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP ME IM IN A BIG HURRY
Alex787 [66]
Part A:
Price before tax: $49.99
Discount: 50% of $49.99 = 50% * $49.99 = 0.50 * $49.99 = $25.00
Discounted price: $49.99 - $25.00 = $24.99
Tax on discounted price: 8% of $24.99 = 8% * $24.99 = 0.08 * $24.99 = $2.00

Part B:
Total price: $24.99 + $2.00 = $26.99
5 0
4 years ago
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