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Solnce55 [7]
3 years ago
6

REPOST PLEASE I NEED HELP I could really use some help with this guys:).........................................................

.......................................................................................................................................................................................Groundhog's Day is on February 2nd each year. On this day, in Punxsatawney, Pennsylvania, thousands of onlookers watch to see if Phil, the groundhog, will cast a shadow. If he does, then six more weeks of winter are predicted. If he doesn't, then an early spring is predicted. Historically, Phil's prediction has only been correct about 40% of the time.
Using complete sentences, design a simulation to determine the probability that Phil's prediction will be correct at least two out of the next three years. Make sure to explain how you would use the simulation to determine the probability.
Mathematics
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

Points

Step-by-step explanation:

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Will give brainliest, thanks in advance!!
tester [92]

Answer:

-3+15i

Step-by-step explanation:

Apply complex arithmetic rule: \left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i

\left(-3\cdot \:3-3\left(-2\right)\right)+\left(-3\left(-2\right)+3\cdot \:3\right)i

-3 * 3 -3 (-2) = -3

-3+15i

8 0
2 years ago
What would the correct answer be???
tiny-mole [99]

Answer:

8.27837797

I think this is correct im not completely sure.

8 0
3 years ago
Vong grilled 21 burgers. He grilled the same number of pounds of turkey burgers as burgers. Turkey burger weighed 1/4 pound and
d1i1m1o1n [39]

Answer:

12 turkey burgers 9 hamburgers

Step-by-step explanation:

5 0
3 years ago
The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a
Vinil7 [7]

Hi, you've asked an incomplete question. However, I inferred you need a brief explanation about The Collatz conjecture.

<u>Explanation:</u>

Put simply, what the Collatz conjecture unsolved problem entails is that if any positive number is picked and it is:

  1. An even number (eg 2, 4, 6,...), then if they are divided by 2,  the new number gotten should undergo the same process (that is to be divided by 2), it is believed your calculation would finally end up at 1. For example, let's pick the number 6, (6÷2=4; repeating the process 4÷2=<u>1</u>)
  2. An odd number, then if they are multiplied by 3 and 1 is added to the result, it is believed that your calculation would finally end up at 1.
6 0
2 years ago
In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g
KIM [24]

Answer:

3.52 \times 10^{-9} = 3.52 \times 10^{-7}\% probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

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)!}

All girls from the first group:

20 students, so N = 20

10 girls, so k = 10

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163

All girls from the second group:

20 students, so N = 20

5 girls, so k = 5

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006

All girls from the third group:

20 students, so N = 20

8 girls, so k = 8

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}

3.52 \times 10^{-9} = 3.52 \times 10^{-7}\% probability that all the 15 students selected are girls

7 0
3 years ago
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