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3 years ago

BRAINLIESTTTTTT ASAPPPP PLEASE ANSWER MY QUESTION

Mathematics

1 answer:

yawa3891 [41]3 years ago

7 0

The red line is represented by y ≥ -3x + 3.

The blue line is represented by y < 3/2x - 6.

A) In this system of inequalities, there is one solid line and one dashed line.

The solid line is the inequality y ≥ -3x + 3 because it has the symbol greater than or equal to. The "equal to" part of the symbol is important because it means that any coordinate located on the solid line is a solution.

However, the inequality y < 3/2x - 6 is a dashed line because it does not have "equal to." This makes the inequality a dashed line which means that all coordinates on it will not be a solution to the inequality.

In the inequality y ≥ -3x + 3, the shading is above the line. This means that all coordinates within the line and shaded area above the line will be a solution to the inequality.

In the inequality y < 3/2x - 6, the shading is below the line. This means that all coordinates within the shared area of the line will be a solution to the inequality. However, since y is not less than or EQUAL TO, all coordinates within the dashed line is not a solution to the inequality.

The solution area, where a coordinate is a solution to
both inequalities, is where the shaded areas of both inequalities overlap.

B) The point (-6, 3) is a not solution to both of the inequalities because it is not within the shared areas of the two inequalities.

Substitute the coordinate (-6, 3) in the inequality y < 3/2x - 6.

y < 3/2x - 6

3 < 3/2(-6) - 6

3 < -9 - 6

3 < -15

This is not true because a negative number cannot be greater than a positive number. Therefore, this proves that the coordinate (-6, 3) is not a solution for the inequality y < 3/2x - 6.

Check if the coordinate fits for the other inequality.

y ≥ -3x + 3

3 ≥ -3(-6) + 3

3 ≥ 18 + 3

3 ≥ 21

This is also not true because 3 is not greater than or equal to 21. Therefore, this coordinate is not a solution to the inequality y ≥ -3x + 3.

The coordinate (-6, 3) is not a solution for the system of inequalities because it is not a solution for both inequalities.

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a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

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2 years ago