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alexandr1967 [171]
2 years ago
3

Use the graph and included equation to answer the question.

Mathematics
1 answer:
Scilla [17]2 years ago
8 0

Using the given linear function, it is found that:

The slope indicates that the average weight of an apple is of 0.32 pounds and the y-intercept represents the weight of the basket is of 1.46 pounds when it is empty.

<h3>What is a linear function?</h3>

A linear function is modeled by:

y = ax + b

In which:

  • a is the slope, which is the rate of change, that is, by how much y changes when x changes by 1.
  • b is the y-intercept, which is the value of y when x = 0, and can also be interpreted as the initial value of the function.

In this problem, the function is:

y = 0.32x + 1.46.

Hence:

The slope indicates that the average weight of an apple is of 0.32 pounds and the y-intercept represents the weight of the basket is of 1.46 pounds when it is empty.

More can be learned about linear functions at brainly.com/question/24808124

#SPJ1

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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

So

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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