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3 years ago

The points on the graph show how much Chef Allen pays for different amounts of beans

Mathematics

2 answers:

Nitella [24]3 years ago

8 0

Answer:

Step-by-step explanation:

Leni [432]3 years ago

7 0

Answer:

See explanation

Step-by-step explanation:

The ordered pair (25,20) shows that Chef Allen pays $20 for 20 pounds of beans. The ratio of pounds to dollars is

$\dfrac{25}{20}=\dfrac{5}{4}$

A. Another plotted points are (5,4), (10,8), (15,12) and (20,16). Find the ratio of pounds to dollars in each case:

$(5,4):\ \ \ \dfrac{5}{4}\\ \\(10,8):\ \ \ \dfrac{10}{8}=\dfrac{5}{4}\\ \\(15,12):\ \ \ \dfrac{15}{12}=\dfrac{5}{4}\\ \\(20,16):\ \ \ \dfrac{20}{16}=\dfrac{5}{4}$

All points show the same ratio.

B. The line connecting given points is the graph showing how much Chef Allen pays for different amounts of beans. This line starts at the origin (0,0) and passes through the points (5,4), (10,8), (15,12) and (20,16).

C. If he cost of 25 pounds of beans is $20, then the cost of 50 pounds of beans is $40.

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a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you

aniked [119]

Answer:

0.64 = 64% probability that the student passes both subjects.

0.86 = 86% probability that the student passes at least one of the two subjects

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Passing subject A

Event B: Passing subject B

The probability of passing subject A is 0.8.

This means that $P(A) = 0.8$

If you have passed subject A, the probability of passing subject B is 0.8.

This means that $P(B|A) = 0.8$

Find the probability that the student passes both subjects?

This is $P(A \cap B)$ . So

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

$P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64$

0.64 = 64% probability that the student passes both subjects.

Find the probability that the student passes at least one of the two subjects

This is:

$p = P(A) + P(B) - P(A \cap B)$

Considering $P(B) = 0.7$ , we have that:

$p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86$

0.86 = 86% probability that the student passes at least one of the two subjects

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3 years ago

The National Safety Council publishes information about automobile accidents in Accident Facts. According to that document, the

vladimir1956 [14]

Answer:

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$

In which x is the number of which we want to find the probability.

Step-by-step explanation:

For each traffic fatality, there are only two possible outcomes. EIther it involved an intoxicated or alcohol-impaired driver or nonoccupant, or it didn't. Traffic fatalities are independent of other traffic fatalities, which means that 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.

The probability is .40 that a traffic fatality involves an intoxication or alcohol-impaired driver or nonoccupant.

This means that $p = 0.4$

Eight traffic fatalities

This means that $n = 8$

Find the probability that the number which involve an intoxicated or alcohol-impaired driver or nonoccupant is

This is P(X = x), in which x is the number of which we want to find the probability. So

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

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$

3 0

3 years ago

A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats

MrRa [10]

Answer:

The three statements are true.

Step-by-step explanation:

The question is incomplete:

A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.

A. There are 54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

C. There are 16 clients who own cats but not dogs.

A. There are 54 clients who own dogs.

TRUE. Of the 70 clients, only 36 own cats. There are left 34 clients that own only dogs. If we add the 20 clients that own both cats and dogs, we have 34+20=54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

TRUE. Of the 70 clients, only 36 own cats. Then, there are left 70-36=34 clients that own only dogs.

C. There are 16 clients who own cats but not dogs.

TRUE. Out of the 36 clients that own cats (only of with dogs), there are 20 that own both. Therefore, there are 36-20=16 clients that own only cats.

5 0

3 years ago