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3 years ago

Please please please help I’ll give you brainliest

Mathematics

2 answers:

Oduvanchick [21]3 years ago

8 0

Find a common factor between 4 and itself to which is 4. So you can then divide the ratio by 4 on both sides to get 1:1

Levart [38]3 years ago

5 0

Answer:

1, 1:1

Step-by-step explanation:

Just take 4:4 ÷ 4 to get 1:1

You might be interested in

Suppose that 35 people are divided in a random mannerinto two teams in such a way that one team contains10 people and the other

Alisiya [41]

To be precise, the size of your sample space is (2410)(2410). This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, (22)(22) (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are 24−2=2224−2=22 nondefective bulbs, and I must choose 88 of them, so that's (228)(228). Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: (22)(228)(22)(228)

P=(22)(228)(2410)P=(22)(228)(2410)

Or, since (22)=1(22)=1,

P=(228)(2410)P=(228)(2410)

Hope this helps!

7 0

3 years ago

Which fact is related to 12 : 3 = 4

goblinko [34]

Answer:

3x4 = 12 is the most related fact to 12:3 = 4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

How many true conditional statements may be written using the following statements?

tangare [24]

The answer is A, Hope this helps <3... can i have brainlyest?

4 0

3 years ago

Determine the number of ways a computer can randomly generate one or more such integers from 1 through 30. Two distinct integers

Thepotemich [5.8K]

Answer:

There are 26 possible way to determine two distinct integers whose sum is 27.

Step-by-step explanation:

To find : The number of ways a computer can randomly generate one or more such integers from 1 through 30. Two distinct integers whose sum is 27.

Solution :

We have given the numbers from 1,2,3,4......,29,30.

In order to get two distinct numbers having the sum 27,

There are the possibilities :

1+26=27

2+25=27

3+24=27

......

24+3=27

25+2=27

26+1=27

The maximum number taken is 26.

So, There are 26 possible way to determine two distinct integers whose sum is 27.

5 0

3 years ago