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Paladinen [302]
3 years ago
13

Pat picked a card from a standard deck, looked at it, and then put it back. He then picked a second card. What is the probabilit

y that Pat pick a diamond or a jack on either pick
Mathematics
1 answer:
AlekseyPX3 years ago
3 0

Answer: 17/52

Step-by-step explanation:

There are 52 total cards in a standard deck.

The number of diamonds in a deck of cards is 13 and there are 4 jacks in a deck of cards.

So the propabibilty of Pat picking a diamond is 13/52 and the probability of Pat picking a jack is 4/52. Now you have to add the two probabilities together.

You do not remove any digits because Pat put the card back into the deck, he didn't take it away. So we do not take away from the total

13 + 4 = 17

You do not need to change the denominator cause 52 is the same.

So 17/52 is the probability of Pat picking a diamond or a jack

I hope this helped!

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If 10 candy cost $35. How much candy can I buy with $20
Alika [10]

Answer:

6 candies

Step-by-step explanation:

(find unit rate, cost / quantity)

35/10 = 3.5

20/3.5 = = 5.7142857142857

candies cannot be in decimal so you must round to the nearest whole

8 0
2 years ago
At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = \dfrac{15}{120}=0.125

As we know the poisson process, we get that

P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

So, for exactly one car would be

P(n=1) is given by

=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

7 0
3 years ago
A sequence is defined by the following:
katrin2010 [14]

Answer:

Step-by-step explanation:

a1 = 6

a2 = -1.2*a_(n-1)

a2 = -1.2*a_1

a2 = -1.2 * 6

a2 = -7.2

a3 = -1.2*a_(n-1)

a3 = -1.2*a_2

a3 = -1.2 * - 7.2

a3 = 8.64

a4 = -1.2 * a_3

a4 = -1.2 * 8.64

a4 = -10.368

6 0
3 years ago
Read 2 more answers
A certain forest covers an area of 4600km. Suppose that each year this area decreases by 7.75%. What will the area be after 13 y
wolverine [178]

Answer:

1679.60\ km^2

Step-by-step explanation:

-We can use the exponential decay function to estimate the size after 13 years:

P_t=P_oe^{-rt}

Where:

P_t is the size after t years, t is the time, r is the rate of decay and P_o the original size.

#We substitute and calculate as:

P_t=P_oe^{-rt}\\\\=4600e^{-13\times 0.0775}\\\\=1679.60

Hence, the forest covers 1679.60\ km^2 after 13 years.

4 0
3 years ago
Identify the type of function represented by
SCORPION-xisa [38]
That's definitely an example of exponential decay, since the base (1/2) (also called the "common ratio") is greater than 0 but less than 1.
6 0
3 years ago
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