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frosja888 [35]
4 years ago
9

A company has developed a wristband for monitoring blood sugar levels without requiring direct blood samples. It is interested i

n demonstrating the accuracy of the device for governmental approval and has decided to test the claim "The glucose level reported by the wristband is within 10% of a standard blood test result." Which of the following data collection processes would be appropriate? Select only one answer choice.
A. Choose a random sample of employees at lunchtime and measure their blodd sugar using both the wristband and a standard blood test.
B. Choose a random sample of people from the local area and random times throughout the day and measure their blood sugar using both the wristband and a standard blood test.
C. Choose a random sample of people from the local area and random times and measure their blood sugar using the wristband. Choose another random sample of people and random times and measure their blood sugar using the standard blood test.
D. Go to a hospital and have the doctors choose a random sample f patients to be tested at random times using both the wristband and the standard blood test.
Mathematics
1 answer:
djverab [1.8K]4 years ago
5 0

Answer:

B. Choose a random sample of people from the local area and random times throughout the day and measure their blood sugar using both the wristband and standard blood test.

Step-by-step explanation:

This is because here we are using a sample that is not limited to one time(like lunch time as option 1), the blood reading are of the same person(unlike option 3) and is not biased due to selective sample collection(like patients in a hospital who normally will show a greater range of results)

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bogdanovich [222]

Answer:

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We can express these relationships algebraically as well as graphically.

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3 years ago
The scale of a model airplane is 1:72.
ExtremeBDS [4]
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3 years ago
A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90​% of the time when t
garik1379 [7]

Answer:

a)  P[A/B] = 0,019     or     P[A/B] = 1,9 %

b)  P[A- /B-] = 0,9996       or    P[A- /B-] = 99,96 %

Step-by-step explanation:

Bayes Theorem :

P[A/B]  =  P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1        real infection     1/300        and     not infection  299/300

Then

1.-    1/300      299/300

When the test is done   (virus present)  0,9 (+)    0,15 (-)

2.-   299/300

When the test is done  ( no virus )   0,15  (+)     0,85 (-)

Then:

P(A) = event person infected          P(B)  =  person test positive

a) P[A/B]  = P(A) * P[B/A] / P(B)

where   P(A)  = 1/300  =   0,0033   P[B/A] = 0,9    

Then P(A) * P[B/A] =  0,0033*0,9  =  0,00297

P(B)   is    ( 1/300 )*0,9  +  (299/300)*0,15

P(B) = 0,0033*0,9 + 0,9966*0,15    ⇒  P(B) = 0,1524

Finally

P[A/B] =  0,00297 /0,1524

P[A/B] = 0,019     or     P[A/B] = 1,9 %

b) Following sames steps:

P[A- /B-] = (299/300) * 0,85  / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

P[A- /B-] = 0,9996       or    P[A- /B-] = 99,96 %

6 0
3 years ago
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Vlad [161]

Answer:

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Please help me!
OLEGan [10]
You need to find 7 1/2% of $29.95.
To do that, just multiply the percent by the number.

7 1/2% * $29.95 =

= 7.5% * $29.95

= 0.075 * $29.95

= $2.24625

The answer is $2.25
7 0
4 years ago
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