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lisov135 [29]
3 years ago
6

Which system of linear inequalities is graphed? {y>2x+1x+y<−2 {y<2x+1x+y>−2 {y≤2x+1x+y≥−2 {y≥2x+1x+y≤−2 The image is

a system of linear inequalities graphed on a coordinate plane with increments of 1 and x and y axis ranging from negative 5 to 5. A dashed line passes through the points begin ordered pair 0 comma 1 end ordered pair and begin ordered pair 1 comma three end ordered pair. The shading is above the line. The other line passes through begin ordered pair 0 comma negative 2 end ordered pair and begin ordered pair negative 2 comma zero end ordered pair. The shading is below this line.

Mathematics
1 answer:
shepuryov [24]3 years ago
4 0

Let us check the slope of y-intercepts of the given lines in the graph first.

The y-intercept of above line is 1 and slope of

Rise/run = 2/1 (Moving 2 units up and 1 unit right).

So, the equation should be

y=2x+1

Y-intercept of second line is -2 and slope is

Rise/run = -1/1 (Moving 1 unit down and 1 unit right).

So, the equation should be

y=-x-2.

Now, we need to check the shaded portion for inequality signs.

We have both lines dotted.

So, the inequality signs would be just < or >.

For y=2x+1 line : Shading is on the left side.

On the left side of the line y=2x+1, the y-values are greater than on right side.

Therefore, we got first inequality y>2x+1

For y=-x-2 line : Shading is on the down of the line.

On the down of the line y=-x-2, the y-values are less than up side of the line.

Therefore, second inequality would be

y<-x-2 or y+x<-2

Therfore, correct option is first option y>2x+1 and y+x<-2.


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Answer:

(a) Pr = 0.3024

(b) Pr = 0.6976

(c) Pr = \frac{^9P_{n-1}}{10^{n-1}}

Step-by-step explanation:

Given

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n(Start) =4

Digits = 5

Solving (a): Probability that each of the 5 digit are different

Since there is no restriction;

The total possible selection is as follows:

First\ digit = 4 (i.e. any of the 4 start digits)

Second\ digit = 10\\ (i.e. any of the 10 digits 0 - 9)

Third\ digit = 10 (i.e. any of the 10 digits 0 - 9)

Fourth\ digit = 10 (i.e. any of the 10 digits 0 - 9)

Fifth\ digit = 10 (i.e. any of the 10 digits 0 - 9)

So, the total is:

Total = 4 * 10 * 10 * 10 * 10

Total = 40000

For selection that all digits are different, the selection is:

First\ digit = 4 (i.e. any of the 4 start digits)

Second\ digit = 9 (i.e. any of the remaining 9)

Third\ digit = 8 (i.e. any of the remaining 8)

Fourth\ digit = 7 (i.e. any of the remaining 7)

Fifth\ digit = 6 (i.e. any of the remaining 6)

So:

Selection =4 * 9 * 8 * 7 * 6

Selection =12096

So, the probability is:

Pr = \frac{Selection}{Total}

Pr = \frac{12096}{40000}

Pr = 0.3024

Solving (b): At least 1 repeated digit

The probability calculated in (a) is the all digits are different i.e. P(None)

So, using laws of complement

We have:

P(At\ least\ 1) = 1 - P(None)

So, we have:

Pr= 1 - 0.3024

Pr = 0.6976

Solving (c): An expression to model the probability.

<em>Using (a) as a point of reference, we have;</em>

Pr = \frac{Selection}{Total}

Where

Selection =4 * 9 * 8 * 7 * 6 ---- for selection of 5 i.e. n = 5

Total = 4 * 10 * 10 * 10 * 10

Selection =4 * 9 * 8 * 7 * 6

This can be rewritten as:

Selection = 4 * ^9P_4

4 can be expressed as: 5 - 1

So, we have:

Selection = (5-1) *^9P_{5-1}

Substitute n for 5

Selection = (n-1) *^9P_{n-1}

Selection = (n-1)^9P_{n-1}

Total = 4 * 10 * 10 * 10 * 10

This can be rewritten as:

Total = 4 * 10^4

Total = (5-1) * 10^{5-1}

Total = (n-1) * 10^{n-1}

Total = (n-1) 10^{n-1}

So, the expression is:

Pr = \frac{(n-1)^9P_{n-1}}{(n-1)10^{n-1}}

Pr = \frac{^9P_{n-1}}{10^{n-1}}

<em>Where n represents the digit number</em>

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