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soldi70 [24.7K]

3 years ago

11

In a shipment of 25 computer monitors, 3 monitors havepossible defects. The first monitor tested from the shipment is defective

and it is removed from the shipment. What is the probability that the second monitor tested is also defective?

2 answers:

otez555 [7]3 years ago

8 0

Answer:

there is a 2 out of 24 chance that the second computer is defective

Step-by-step explanation: since the first one is defective that makes only 2 possible defective monitors which also makes 24 computers left because that first defective one was removed

Citrus2011 [14]3 years ago

6 0

It has a 2 in 24 chance of being picked next

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\frac{(2_1-4_{2}) ^{4} } {2} <br> Quick I need help<br> Plz answer it NOW

erastova [34]

Answer:

2(2^1-4^2)^4

Step-by-step explanation:

6 0

3 years ago

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish

Anna007 [38]

Answer:

Ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

Step-by-step explanation:

Let the number of red fish and blue fish in the large tank are x and y respectively.

Similarly ratio of red fish and blue fish in the small tank are x' and y' respectively.

Since in each tank ratio of the red fish to blue fish is 333 : 444

That means x : y = 333 : 444

Or $\frac{x}{y}=\frac{333}{444}$

⇒ $\frac{x}{y}=\frac{3}{4}$

⇒ y = $\frac{4x}{3}$ --------(1)

Similarly x' : y' = 333 : 444

⇒ $\frac{x'}{y'}=\frac{333}{444}$

⇒ $\frac{x'}{y'}=\frac{3}{4}$

⇒ x' = $\frac{3y'}{4}$ ------(2)

Ratio of the fish in large tank to the fish in small tank is 464646 : 555

So (x + y) : (x' + y') = 464646 : 555

$\frac{x+y}{x'+y'}=\frac{464646}{555}$

Now we replace the values of x and y' from equation (1) and equation (2)

$\frac{(x+\frac{4x}{3})}{(\frac{3y'}{4}+y')}=\frac{464646}{555}$

$\frac{\frac{7x}{3}}{\frac{7y'}{4}}=\frac{464646}{555}$

$\frac{4}{3}\times \frac{x}{y'}=\frac{464646}{555}$

$\frac{x}{y'}=\frac{3}{4}\times \frac{464646}{555}$

$\frac{x}{y'}=\frac{232323}{370}$

$\frac{y'}{x}=\frac{370}{232323}$

$\frac{y'}{x}=\frac{10}{6279}$

Therefore, ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

4 0

3 years ago

Which statement is correct?

MakcuM [25]

Answer:

the third 1

Step-by-step explanation:

3 0

3 years ago

the late fee for library books is $2.00 plus 15 cents each day for a book that is late. If Maria's fee for a late book was $3.20

love history [14]

Answer:

Maria's book was 8 days late

Step-by-step explanation:

According to the information given, you can write an equation that states that 15¢ multiply for the number of days that the book was late plus $2 is equal to $3.20:

0.15x+2= 3.20

x= number of days that the book was late

Now, you have to solve for x:

0.15x= 3.20-2

0.15x= 1.20

x= 1.20/0.15

x= 8

The book was 8 days late.

5 0

2 years ago

Assume that when adults with smartphones are randomlyâ€‹ selected, 63â€‹% use them in meetings or classes. If 7 adult smartphone

nlexa [21]

Answer:

5.78% probability that exactly 2 of them use their smartphones in meetings or classes.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they use their smarthphone in meetings or classes, or they do not. The probability of an adult using their smartphone on meetings or classes is independent of other adults. So 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.

63% use them in meetings or classes.

This means that $p = 0.63$

7 adult smartphone users are randomly selected

This means that $n = 7$

Find the probability that exactly 2 of them use their smartphones in meetings or classes.

This is P(X = 2).

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

$P(X = 2) = C_{7,2}.(0.63)^{2}.(0.37)^{5} = 0.0578$

5.78% probability that exactly 2 of them use their smartphones in meetings or classes.

8 0

3 years ago