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sasho [114]
3 years ago
15

Please helppp!! I don't know how to do this

Mathematics
2 answers:
Lera25 [3.4K]3 years ago
7 0

Answer:

it would be (-2,-4)

mylen [45]3 years ago
4 0

Answer:

I = (- 2, - 4 )

Step-by-step explanation:

point I is positioned 2 left ( - direction ) and 4 down ( - direction ) from O

The ordered pair ( coordinates ) of point I = (- 2, - 4 )

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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
Math question down below
Dafna11 [192]
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3 0
3 years ago
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A survey claims that a college graduate from Smith College can expect an average starting salary of $ 42,000. Fifteen Smith Coll
jek_recluse [69]
<h2>Answer with explanation:</h2>

Let \mu be the average starting salary ( in dollars).

As per given , we have

H_0: \mu=42000\\\\ H_a:\mu

Since H_a is left-tailed , so our test is a left-tailed test.

WE assume that the starting salary follows normal distribution .

Since population standard deviation is unknown and sample size is small so we use t-test.

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}} , where n= sample size , \overline{x} = sample mean , s = sample standard deviation.

Here , n= 15 , \overline{x}=  40,800  , s= 225

Then, t=\dfrac{40800-42000}{\dfrac{2250}{\sqrt{15}}}\approx-2.07

Degree of freedom = n-1=14

The critical t-value for significance level α  = 0.01 and degree of freedom 14 is 2.62.

Decision : Since the absolute calculated t-value (2.07) is less than the  critical t-value., so we cannot reject the null hypothesis.

Conclusion : We do not have sufficient evidence at 1 % level of significance to support the claim that  the average starting salary of the graduates is significantly less that $42,000.

3 0
3 years ago
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