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stepladder [879]
3 years ago
12

Gru's schemes have a/an 11% chance of succeeding. An agent of the Anti-Villain League obtains access to a simple random sample o

f 1100 of Gru's upcoming schemes.Find the probability that...(Answers should be to four places after the decimal, using chart method, do NOT use the continuity correction):

Mathematics
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

The complete solution is given in the attachments.

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Temka [501]
4/1 x 3/2 = 12/2
6/9 x 1/7 = 6/63
5/6 x 2/1 = 10/6
8/3 x 1/6 = 8/18
8/1 x 5/8 = 40/8
7 0
2 years ago
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Use the cosine sum and difference identities to find the exact value.<br><br> COS(5pi/12)
nydimaria [60]
5π/12 = (5 · 180°) : 12 = 75°
cos 75° = cos ( 45° + 30° )= cos 45° cos 30° - sin 45° sin 30° =
=√2/2 * √3/2 - √2/2 * 1/2 = \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4}= \frac{ \sqrt{6} - \sqrt{2} }{4}
=(2.4495 - 1.4142): 4 = 0.258825
8 0
2 years ago
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The prostate-specific antigen (PSA) test is a simple blood test to screen for prostate cancer. It has been used in men over 50 a
Ugo [173]

The question is incomplete. I am writing the complete question below:

The prostate-specific antigen (PSA) test is a simple blood test to screen for prostate cancer. It has been used in men over 50 as a routine part of a physical exam, with levels above 4 ng/mL indicating possible prostate cancer. The test result is not always correct, sometimes indicating prostate cancer when it is not present and often missing prostate cancer that is present. Suppose that these are the approximate conditional probabilities of a positive (above 4 ng/ml) and negative test result given cancer is present or absent.

                                    Positive Result                 Negative Result

Cancer Present                   0.21                                       0.79

Cancer Absent                    0.06                                      0.94

In a large study of prostate cancer screening, it was found that about 6.6% of the population has prostate cancer.

What is the probability that the test is positive for a randomly chosen person from this population? (Enter your answer to five decimal places.)

P(Positive test) =

Answer:

P(Positive Result) = 0.0699

Step-by-step explanation:

We are given that 6.6% of the population has prostate cancer. So,

P(Cancer Present) = 0.066

P(Cancer Absent) = 1 - 0.066 = 0.934

From the given conditional probability table, we have:

P(Positive Result | Cancer Present) = 0.21

P(Positive Result | Cancer Absent) = 0.06

We need to find the probability that the result is positive. It can be either that the result is positive and cancer is present or the result is positive and the cancer is absent. So,

P(Positive Result) = P(Positive Result∩Cancer Present) + P(Positive Result∩Cancer Absent)

                             = P(Positive Result | Cancer Present)*P(Cancer Present) + P(Positive Result | Cancer Absent)*P(Cancer Absent)

                             = (0.21)*(0.066) + (0.06)*(0.934)

                             = 0.01386 + 0.05604

P(Positive Result) = 0.0699

5 0
3 years ago
Please help! 5 3/4-3 1/4=?
KonstantinChe [14]
Your answer will be 2 2/4. If you want it simplified it will be 2 1/2
6 0
3 years ago
No links please i cant seem to find the answer for this qn
Rudiy27

Answer:

(a) is 15

(b) is 30

75 divide by 15 times 90 divide by 15 is 30

6 0
2 years ago
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