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olya-2409 [2.1K]
2 years ago
14

Assume that a study of 500 randomly selected airplane routes showed that 482 arrived on time. Select the correct interpretation

of the probability of an airplane arriving late. Interpret an event as significant if its probability is less than or equal to 0.05.
A. Significant at 0.0036
B. Not significant at 0.964
C. Significant at 0.036
D. Not significant at 0.036
Mathematics
1 answer:
Westkost [7]2 years ago
4 0

Answer: C. Significant at 0.036

Step-by-step explanation:

Given:

Number of selected samples Ns= 500

Number of airplane that arrive on time Na = 482.

Number of airplane that arrive late Nl = 500 - 482 = 18

The probability that an airplane arrive late:

P(L) = Nl/Ns

P(L) = 18/500

P(L) = 0.036

Interpret an event as significant if its probability is less than or equal to 0.05.

Since P(L) < 0.05

P(L) = Significant at 0.036

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Can u solve this pls easy? A geologist takes samples of two rocks from a mountainside. The gray rock has a volume 25% higher tha
Korolek [52]

Answer:

Text me back for an answer.

Step-by-step explanation:

Make sure you text back.

5 0
2 years ago
Conservation of Species A certain species of turtle faces extinction because dealers collect truckloads of turtle eggs to be sol
JulijaS [17]

Answer:

The turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

Ten years after the conservation measures are implemented the population will be 3260 turtles.

Step-by-step explanation:

To find the rate of growth of the turtle population at any time <em>t</em> you need to find N'(t)

\frac{d}{dt}N(t)=\frac{d}{dt}(2t^3+3t^2-4t+1000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dt}(2t^3)+\frac{d}{dt}(3t^2)-\frac{d}{dt}(4t)+\frac{d}{dt}(1000)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\N'(t)=6t^2+6t-4

In particular, when t = 2 and t = 6, we have

N'(2)=6(2)^2+6(2)-4=32\\\\N'(6)=6(6)^2+6(6)-4=248

so the turtle population's rate of growth will be 32 turtles per year after 2 years and 248 per year after 6 years.

The turtle population at the end of the tenth year will be

N(10)=2(10)^3+3(10)^2-4(10)+1000\\N(10)=3260 \:turtles

3 0
2 years ago
NEED HELP ON NUMBER 13 AND 14
Kazeer [188]

Answer:

Question 13: For age groups y=1 and y=1.3 response is 8 microseconds.

Question 14: The club was making a loss between 11.28 and 4.88 years.

Step-by-step explanation:

Question 13:

The age group y for which the response rate R is 8 microseconds  is given by the solution of the equation

8=y^4 +2y^3 - 4y^2 -5y +14.

We graph this equation and find the solutions to be

y=1;   y=1.302;     y=-2;    y=-2.302.

Since only positive solutions for y are valid in the real world we take only those.

Thus only for age groups y=1 and y=1.3 the response is 8 microseconds.

Question 14:

The footbal club is making a loss when p(t)

Or

t^3 -14t^2 +20t +120

We graph this inequality and find the solutions to be

t and 4.88

Since in the real world only positive values for t are valid, we take the the second solution to be true.

Thus the club was making a loss in years 4.88

5 0
2 years ago
Simplify this expression.<br><br> 8.9 – 1.4x + (–6.5x) + 3.4
Svetlanka [38]

Answer:

-7.9x+12.3

Step-by-step explanation:

8.9 – 1.4x + (–6.5x) + 3.4

8.9-1.4x-6.5x+3.4

8.9-7.9x+3.4

-7.9x+8.9+3.4

-7.9x+12.3

4 0
2 years ago
Please help me with this question
natulia [17]

what's up

Seek understand, to be understood!

To get the answer we need the formula or source.

The question in another way if 4cm=1.2km, what is the 10cm=--------

first, you have to use the crisscross.

4cm=1.2km

10cm= x it is variable  

then, 4xx= 1.2kmx10cm

4x=12km

4x/4= 12/4

x=3km  


4 0
2 years ago
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