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viva [34]
3 years ago
5

A construction zone on Interstate 15 has a speed limit of 40 mph. The speeds of vehicles passing through this construction zone

are normally distributed with a mean of 44 mph and a standard deviation of 3 mph.
Required:
a. Find the percentage of vehicles who pass through this construction zone who are exceeding the posted speed limit.
b. What percentage of vehicles travel through this construction zone with speeds between 50 mph and 55 mph?
Mathematics
1 answer:
monitta3 years ago
3 0

Answer:

a. The percentage of vehicles who pass through this construction zone who are exceeding the posted speed limit =90.82%

b. Percentage of vehicles travel through this construction zone with speeds between 50 mph and 55 mph= 2.28%

Step-by-step explanation:

We have to find

a) P(X>40)= 1- P(x=40)

Using the z statistic

Here

x= 40 mph

u= 44mph

σ= 3 mph

z=(40-44)/3=-1.33

From the  z-table   -1.67 = 0.9082

a) P(X>40)=

Probability exceeding the speed limit = 0.9082 = 90.82%

b) P(50<X<55)

Now

z1 = (50-44)/3 = 2

z2 = (55-44)/3= 3.67

Area for z>3.59 is almost equal to 1

From the z- table we get

P(55 < X < 60) = P((50-44)/3 < z < (55-44)/3)

     = P(2 < z < 3.67)

     = P(z<3.67) - P(z<2)

     = 1 - 0.9772

     = 0.0228

or 2.28%

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7 0
3 years ago
Simplify 2x – 9y + 4x + 11y by combining like terms and then evaluate at x = -2<br> and y = 4.
Mashcka [7]

Step-by-step explanation:

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First of all combining like terms

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3 years ago
What’s the volume of a rectangle solid with the given dimensions: length 20, width 23, height 26
ANEK [815]

The volume of the rectangular solid is: 11,960 units³.

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8 0
2 years ago
100 points!!!
Bogdan [553]

Answer:

(1, 5)

Step-by-step explanation:

The solution to the system of equations is the point of intersection of the two lines.  From inspection of the graph, the point of intersection is at (1, 5).

<u>Proof</u>

The solution to a system of equations is the point at which the two lines meet.  

⇒ g(x) = f(x)

⇒ 3x + 2 = |x - 4| + 2

⇒ 3x = |x - 4|

⇒ 3x = x - 4   and   3x = -(x - 4)

⇒ 3x = x - 4

⇒ 2x = -4

⇒ x = -2

Inputting x = -2 into the 2 equations:

⇒ g(-2) = 3 · -2 + 2 = -4

⇒ f(-2) = |-2 - 4| + 2 = 8

Therefore, as the y-values are different, x = -2 is NOT a solution

⇒ 3x = -(x - 4)

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Inputting x = 1 into the 2 equations:

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Therefore, as the y-values are the same, x = 1  IS a solution

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8 0
3 years ago
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jekas [21]

Answer:

C

Step-by-step explanation:

7 0
3 years ago
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