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3 years ago

8

Lella will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $46 and costs an ad

ditional $0.14 per mile driven. The
second plan has an initial fee of $51 and costs an additional $0.10 per mile driven

1 answer:

Helen [10]3 years ago

6 0

The plans cost the same at 125 miles. the cost is $63.50

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

monitta

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%

3 0

3 years ago

Solve f 10/f = 36/90 f = ?

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

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3 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

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3 years ago

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You are given a problem that involves multiplying a positive number and a negative number.

kenny6666 [7]

Answer:

-3 and -2

Step-by-step explanation:

A better explanation...

(positive)(positive)= positive

(positive)(negative)= negative

(negative)(negative)= positive

(negative)(positive)= negative

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3 years ago

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What is -0.009 to the second power

kondaur [170]

I believe it's -0.000081

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3 years ago