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Neko [114]
3 years ago
12

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 81 mph and a standa

rd deviation of 8 mph. The speed limit is 65. If you pick a car on the highway at random, what is the probability the vehicles is going less than or equal to the speed limit?

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
3 0

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: speed of a vehicle along a stretch of I-10 (mph)

This variable has a normal distribution with mean μ= 81 mph and a standard deviation σ= 8 mph.

The speed limit in the said stretch is 65 mph.

You need to calculate the probability of picking a car at random and its speed be at most 65 mph, symbolically:

P(X≤65)

To reach the probability, you need to use the standard normal distribution. To standardize the value fo X you have to subtract the value of μ and then divide it by σ:

P(Z≤(65-81)/8)= P(Z≤-2.00)

Now you look for the corresponding probability in the table of the standard normal distribution, since the value is negative you have to use the left entry. The integer and first decimal numbers are in the first column and the second decimal number is in the first row.

P(Z≤-2.00)= 0.0228

I hope it helps!

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What is the solution?<br> A. 1.87<br> B. 1.61<br> C. 3.73<br> D. 1.13
quester [9]
Try A:
6^(2*1.87) = 813.365

Try B:
6^(2*1.61) = 320.366

Try C:
6^(2*3.73) = 638275.725

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3 years ago
Can anyone help me with this math question
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Answer:

89 seats are not being used

Step-by-step explanation:

You subtract the seats being used by teachers

342 - 75 =267

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3 years ago
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Evgesh-ka [11]
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22.5 or 5/22 or 30

Step-by-step Explanation:

-2 and -2
SSS
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22.5 or 5/22

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7 0
3 years ago
A person x inches tall has a pulse rate of y beats per​ minute, as given approximately by yequals600 x Superscript negative 1 di
tankabanditka [31]

Answer:

a) 5.13beats/min

b) 2.82 beats/min

Step-by-step explanation:

Given the pulse rate of a person modelled by the equation y = 600x^-1/3 for 30≤x≤75

If the height is 39inches, the instantaneous rate of change of pulse rate for the heights will be expressed as;

y = 600(39)^-1/3

y = {600(1/39)}/3

y = 600/39×3

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y ≈ 5.13beats/min

The instantaneous rate for a 39 inches tall person is 5.13 beats per min

b) For a 71inches tall person, the beat rate will be expressed as;

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y = {600(1/71)}/3

y = 600/71×3

y = 600/213

y ≈ 2.82 beats per minute

The instantaneous rate for a 71 inches tall person is 2.82 beats per min

7 0
3 years ago
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