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3 years ago

Match the pairs of variables with the type of relationship they show

Mathematics

2 answers:

kumpel [21]3 years ago

6 0

I believe that correlation without causation is the number of doctors

Correlation with causation is the number of high school seniors. Hope this helps!

Pavlova-9 [17]3 years ago

3 0

Answer:

Correlation without Causation = The number of high school seniors and high school graduates.

Correlation with Causation = The amount of snowfall and the numbers of voters.

Step-by-step explanation:

Since there is a large relationship between the number of high school seniors and high school graduates but there is no cause of this relationship. So it has a Correlation without Causation.

Also, there is a large relationship between the amount of snowfall and the numbers of voters since more the snowfall occurs less will be voters gone for voting. Here is a cause of this relationship. So it has a Correlation with Causation.

and, there is not a good relationship between the number of doctors and the number of nurses.

Further, Correlation shows the strength of the relation between two variables and it does not mean that change in one variable is the cause of change in other variables.

And Causation is the occurrence of an event is the result of the occurrence of the second event.

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In an examination 80% students passed. if 15 less have appeared and 10 less have passed, the ratio of passers to failures would

Nina [5.8K]

Answer:

60 students passed, and 75 appeared in examination.

Step-by-step explanation:

Let's say s is the total number of students and p is the number of students who passed.

80% of the students passed, so:

0.8 s = p

If there were 10 less passers, and 15 less students (5 less failures), then the ratio of passers to failures would be 5/1.

(p − 10) / (s − p − 5) = 5 / 1

Simplify the second equation:

p − 10 = 5 (s − p − 5)

p − 10 = 5s − 5p − 25

6p = 5s − 15

Substitute the first equation.

6 (0.8 s) = 5s − 15

4.8 s = 5s − 15

0.2 s = 15

s = 75

p = 0.8 s

p = 60

60 students passed, and 75 appeared in examination.

5 0

3 years ago

Solve R=M + B, for B.

musickatia [10]

Answer:

B = R - M

Step-by-step explanation:

1. R = M + B Subtract M

2. R - M = B

8 0

4 years ago

Plz solve!!!!! I will give 25 points and brainliest!! and plz, no tiny links<br> 1 3/7*2/5

Airida [17]

Answer:

10/7*2/5= 20/35=4/7

5 0

3 years ago

Read 2 more answers

BRAINLIEST !!! please dont answer if you do not know it

postnew [5]

Answer:

The answer is A.

Step-by-step explanation:

There are Four 6's, Four 4's, Seven 5's, One 2, Two 3's, Two 7's, and One 8.

Hope this helps :))

7 0

3 years ago

A bag contains 4 blue marbles and 2 yellow marbles. Two marbles are randomly chosen (the first marble is NOT replaced before dra

VMariaS [17]

Answer:

0.4 = 40% probability that both marbles are blue.

0.0667 = 6.67% probability that both marbles are yellow.

53.33% probability of one blue and then one yellow

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

Step-by-step explanation:

To solve this question, we need to understand conditional probability(for the final question) and the hypergeometric distribution(for the first three, because the balls are chosen without being replaced).

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

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.

What is the probability that both marbles are blue?

4 + 2 = 6 total marbles, which means that $N = 6$

4 blue, which means that $k = 4$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,4) = \frac{C_{4,2}*C_{2,0}}{C_{6,2}} = 0.4$

0.4 = 40% probability that both marbles are blue

What is the probability that both marbles are yellow?

4 + 2 = 6 total marbles, which means that $N = 6$

2 yellow, which means that $k = 2$

Sample of 2, which means that $n = 2$

This is P(X = 2). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 2) = h(2,6,2,2) = \frac{C_{2,2}*C_{4,0}}{C_{6,2}} = 0.0667$

0.0667 = 6.67% probability that both marbles are yellow.

What is the probability of one blue and then one yellow?

Total is 100%.

Can be:

Both blue(40%)

Both yellow(6.67%)

One blue and one yellow(x%). So

40 + 6.67 + x = 100

x = 100 - 46.67

x = 53.33

53.33% probability of one blue and then one yellow.

If you are told that both selected marbles are the same color, what is the probability that both are blue?

Conditional probability.

Event A: Both same color

Event B: Both blue

Probability of both being same color:

Both blue(40%)

Both yellow(6.67%)

This means that $P(A) = 0.4 + 0.0667 = 0.4667$

Probability of both being the same color and blue:

40% probability that both are blue, which means that $P(A \cap B) = 0.4$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.4667} = 0.8571$

If you are told that both selected marbles are the same color, 0.8571 = 85.71% probability that both are blue

8 0

3 years ago