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lesya [120]
3 years ago
5

What does p equal? -4p=3p 28

Mathematics
1 answer:
Margarita [4]3 years ago
6 0
So first of what sign is in between 3p and 28
You might be interested in
Step by step please and thank you.
cupoosta [38]

Answer:

  the probability is 2/9

Step-by-step explanation:

Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).

  p(dime first) = 4/10 = 2/5

Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.

The joint probability of these two events occurring sequentially is the product of their probabilities:

  p(dime then nickel) = (2/5)×(5/9) = 2/9

_____

<em>Alternate solution</em>

You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.

  p(dime/nickel) = 20/90 = 2/9

8 0
3 years ago
A box contains different colored paper clips. The probability of drawing two red paper clips from the box without replacement is
ivann1987 [24]

Answer: 5/14 which is choice B

================================================

How I got this answer:

Define the following events

A = event of picking a red paper clip on the first selection

B = event of picking a red paper clip on the second drawing

Replacement is not made.

Now onto the probabilities for each

P(A) = 2/5 = 0.4 is given to us as this is simply the probability of picking red on the first try

P(A and B) = probability of both events A and B happeing simultaneously = 1/7

P(B|A) = probability event B occurs, given event A has occured

P(B|A) = probability of selecting red on second selection, given first selection is red (no replacement)

P(B|A) = P(A and B)/P(A)

P(B|A) = (1/7) / (2/5)

P(B|A) = (1/7) * (5/2)

P(B|A) = (1*5)/(7*2)

P(B|A) = 5/14

So if event A happens, then the chances of event B happening is 5/14

------------------

A more concrete example:

If we had 15 paperclips, and 6 of them were red, then

P(A) = (# of red)/(# total) = 6/15 = 2/5

P(B|A) = (# of red left)/(# total left) = (6-1)/(15-1) = 5/14

P(A and B) = P(A)*P(B|A) = (2/5)*(5/14) = 10/70 = 1/7

7 0
4 years ago
Read 2 more answers
How to integrate sqrt(1+x^3)?
Karolina [17]
First integrate the entire thing by letting multiplying by 1/2 the entire expression raised to exponent 1/2 - 1. That is,
                          0.5(1 + x³)^-0.5
Then, multiplying this by the derivative of those inside the parentheses. The final answer would be,
                          (0.5)((1 + x³)^-0.5)(3x²)
8 0
3 years ago
7 = 3t - 3 - 2+ + 6. <br><br>Justification: ? property​
Volgvan

Answer:

7 = 3t + 1

t = 6

....................

7 = 3t - 2t + 3

t = 4

4 0
3 years ago
The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per
ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that \mu = 2

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

P(X \leq 5) + P(X > 5) = 1

We want P(X > 5). So

P(X > 5) = 1 - P(X \leq 5)

In which

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353

P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707

P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707

P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804

P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902

P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361

P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824

P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176

1.76% probability that in one hour more than 5 clients arrive

8 0
3 years ago
Read 2 more answers
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