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Ray Of Light [21]
3 years ago
12

In a study of 225 adults, the average heart rate was 72 beats per minute. Assume the population of heart rates is known to be ap

proximately normal, with a standard
deviation of 10 beats per minute. What does a margin of error of 1.1 for the 90% confidence interval of the average beats per minute mean?
Mathematics
2 answers:
irinina [24]3 years ago
6 0

Answer:

it means that if the survey is repeated there is a 90% chance that the average heart rate is 72 bears per minute with plus or minus 1.1 %

Step-by-step explanation:

Confidence interval explains to what percentage will the result be true if the experiment is repeated multiple times

Margin of error tells the range of accuracy if the result is true when the experiment is repeated multiple times.

Inessa [10]3 years ago
5 0
I think it’s like the root
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B i think that because there’s no quantity
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2 years ago
2/7 into percents ??
mafiozo [28]

Answer:

28.57142857 %

Step-by-step explanation:

Convert the fraction to a decimal, then multiply by  100 .

28.57142857 %

3 0
3 years ago
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Write out the sample space for the given experiment. Use the letter R to indicate red, G to indicate green, and B to indicate bl
Vilka [71]

Answer:

Answer for the question:

Write out the sample space for the given experiment. Use the letter R to indicate red, G to indicate green, and B to indicate blue. A die shows 33 different colors on it. Give the sample space for the next 22 rolls.

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Step-by-step explanation:

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3 0
3 years ago
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
forsale [732]

Answer:

a)

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

Step-by-step explanation:

The formula is:

v=\frac{2 \pi R}{T}

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

a_c=\frac{4 \pi^{2} R}{T^2}

Now,

a)

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}

Centripetal Acceleration at Equator = 3.37*10^{-2} meters per second squared

b)

At 30.0° north of the equator:

R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6

Now,

Speed = v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}

Centripetal Acceleration at 30 degrees north of equator = 2.92*10^{-2} meters per second squared

4 0
3 years ago
3. Solve.<br>-11.0 – 3.5<br>​
Anon25 [30]

Answer:

-14.5

Step-by-step explanation:

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5 0
2 years ago
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