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KATRIN_1 [288]
3 years ago
12

HELP PLEASE!! ASAP WILL MARK BRAINLIEST!! MATH!!

Mathematics
1 answer:
dusya [7]3 years ago
4 0
Its the first one, (-8,8) (2,2)
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HELP PLS! <br> Solve (x-5)^2=3
s344n2d4d5 [400]

Answer:

Step-by-step explanation:Use the square root on both sides:

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7 0
3 years ago
Diego hosted a spaghetti dinner for the soccer team. He made 6 boxes of spaghetti to feed the 20 people that came. Next time, 50
Tems11 [23]
6 boxes of spaghettis for 20 persons so that means 12 boxes for 40 persons since there are 50 persons we have 10 persons left since 10 is equals to 20 divided by 2, divide 6 by 2 which is equals to 3 then add 12
Answer: 15
5 0
3 years ago
The school band wishes to raise a minimum of $800 by selling tickets to their holiday concert. They make $3 from each student ti
aliya0001 [1]
3x+5y>800 minimum is the least 
7 0
3 years ago
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Unknown to a medical researcher, 6 out of 25 patients have a heart problem that will result in death if they receive the test dr
Fiesta28 [93]

Using the hypergeometric distribution, it is found that there is a 0.0002 = 0.02% probability that exactly 6 patients will die.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

The values of the parameters for this problem are:

N = 25, k = 6, n = 8.

The probability that exactly 6 patients will die is P(X = 6), hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 6) = h(6,25,8,6) = \frac{C_{6,6}C_{19,2}}{C_{25,8}} = 0.0002

0.0002 = 0.02% probability that exactly 6 patients will die.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

#SPJ1

8 0
2 years ago
Which quadrant is the following point in?<br> Point A (1,3)
bulgar [2K]
This would be in quadrant 1 on a graph
8 0
3 years ago
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