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nata0808 [166]
3 years ago
14

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. This distribution

takes only whole-number values, so it is certainly not a normal distribution. Let "x-bar" be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate probability that there are fewer than 100 accidents in a year? (Hint: Restate this event in terms of "x-bar")
Mathematics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

The approximate probability that there are fewer than 100 accidents in a year = .9251

Step-by-step explanation:

Given -

Mean (\nu ) =2.2

Standard deviation \sigma = 1.4

Let \overline{X}  be the mean number of accidents per week at the intersection during a year (52 weeks)

Then \overline{X} = \frac{100}{52} = 1.92

the approximate probability that there are fewer than 1.92 accidents per week  in a year

[Z   = \frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}} ]

=  P(\overline{X} < 1.92 )  = (P(\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}< \frac{1.92 - 2.2 }{\frac{1.4}{\sqrt{52}}})

 = P( Z <  -1.442 )   ( Using Z table)

=  .9251

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an object is traveling at a steady speed of 10 1/5 km/h. how long will it take the object to travel 4 4/5
zaharov [31]

Incomplete question but the comnplete question is  how long will it take the object to travel 4 4/5km

Answer: time taken for the object to travel =  0.47 hrs

Step-by-step explanation:

Step 1 -- The formulae to represent the relationship of speed,  distance and time .

Speed = Distance / Time

Where Speed = 10 1/5 km/h

Distance = 4 4/5 km

Step 2 ---  Solving

Time = Distance / Speed

= 4 4/5 km / 10 1/5 km/h

= 24/ 5 x 5/ 51 = 24/51 =0.47 hrs

0r

4. 8 / 10.2 = 0.47 hrs

 Therefore the time taken for the object to travel =  0.47 hrs

6 0
3 years ago
NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, in terms of time is given by h=−4.9t2+244t+269.
Darya [45]

Answer: The height of the rocket after 9 seconds is 2068.1 meters.

The height of the rocket when it initially launched is 269 meters.

Step-by-step explanation:

Given, NASA launches a rocket at t=0 seconds.

Its height, in meters above sea-level, in terms of time is given by h=-4.9t^2+244t+269.

To find, the height of the rocket after 9 seconds put t=9, we get

h=-4.9(9)^2+244(9)+269\\\\=-4.9(81)+2196+269\\\\=-396.9+2465\\\\=2068.1

Hence, the height of the rocket after 9 seconds is 2068.1 meters.

To find, the height of the rocket when it initially launched put t=0, we get

h=-4.9(0)^2+244(0)+269\\\\=0+0+269\\\\=269

Hence, the height of the rocket when it initially launched is 269 meters.

4 0
3 years ago
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Answer:

If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.

Step-by-step explanation:

8 0
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5. Simplify the expression,<br> (-2 - 5i) - (-4+6i)<br> 2 - 11i<br> -2+i<br> -6 + 11i<br> -5i
monitta

\qquad\qquad\huge\underline{{\sf Answer}}♨

Let's simplify ~

\qquad \sf  \dashrightarrow \:( - 2 - 5i) - ( - 4 + 6i)

\qquad \sf  \dashrightarrow \: - 2 - 5i +  4  -  6i)

\qquad \sf  \dashrightarrow \: - 2 +  4   - 5i-  6i)

\qquad \sf  \dashrightarrow \:2 - 11i

Therefore, A is the Correct choice !

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