To make that graph you follow these steps.
1) Set the cartesian coordinate system:
- vertical axis: y
- horizontal axis: x
- positive x-semi axis: to the right
- negative x-semi axis: to the left
- positive y-semi axis: upward
- negative y semi axis: dwonward

2) solve the inequality for y:
given:                   -2x + 5y > 15
transpose-2x:            5y > 15 + 2x
divide by 5:                  y > 3 + (2/5)x

3) Graph

- draw the line y = 3 + (2/5)x, using a dotted line (i.e. - - - - - -)
- remember that 3 is the y intercept, and 2/5 is the slope
- the line is dotted because the solution does not include the points in the line.

- the solution is the region above and to the left of the dotted line.

4) See the figure attached for better visualization: the pink region is the solution of the inequality.

8 0

3 years ago

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Question Help Of 515515 samples of seafood purchased from various kinds of food stores in different regions of a country and gen

Setler79 [48]

Answer:

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

Step-by-step explanation:

Sample size, n = 51

p = 0.62

1 - p = 1 - 0.62 = 0.38

n = 515

Confidence level = 90% = Zcritical at 90% = 1.645

Confidence interval = (p ± margin of error)

Margin of Error = Zcritical * sqrt[(p(1-p))/n]

Margin of Error = 1.645 * sqrt[(0.62(0.38))/515]

Margin of Error = 1.645 * 0.0214

Margin of Error = 0.035203

Lower boundary = (0.62 - 0.035203) = 0.584797

Upper boundary = (0.62 + 0.035203) = 0.655203

(0.5848 ; 0.6552)

We are confident that about 58% to 66% of sea foods in the country are Mislabelled.

No, criticism isnt valid and generalization can be made once the assumptions for constructing a confidence interval is met.

3 0

3 years ago