6. How do you find the slope of a line?

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

2 years ago

PLEASE HELP

Ronch [10]

Answer:

<em><u>1 and 5</u></em>

Step-by-step explanation:

The squares have a side length of 10 and 1 square side is the radius of the half-circles. Since there are two half-circles, find the circumference for one full circle:

$C=2\pi r$

Insert the radius:

$r=10\\C=2\pi *10$

Simplify pi:

$C=2*3.14*10$

Simplify multiplication:

$C=6.28*10\\C=62.8$

The circumference of the circles is 62.8. Now find the perimeter of the exposed squares with side length 10. There are 4 exposed sides, which equals one square. Find the perimeter:

$P=10+10+10+10\\P=10+10+20\\P=10+30\\P=40$

Add the perimeter of the circle and the square together:

$P=62.8+40=102.8$

Now see which of the options gives you the perimeter:

1. $40+20\pi =102.8$ ****

2. $unavailable$

3. $120+20\pi =182.8$

4. $300+100\pi =614.2$

5. $10+10+10\pi +10+10+10\pi =102.8$ ****

Finito.

7 0

2 years ago

1) Given: circle k(O), ED= diameter ,m∠OEF=32°, m(arc)EF=(2x+10)° Find: x

qaws [65]

1. The major arc ED has measure 180 degrees since ED is a diameter of the circle. The measure of arc EF is $(2x+10)^\circ$ , so the measure of arc DF is

$m\widehat{DF}=360^\circ-180^\circ-(2x+10)^\circ=(170-2x)^\circ$

The inscribed angle theorem tells us that the central angle subtended by arc DF, $\angle DOF$ , has a measure of twice the measure of the inscribed angle DEF (which is the same angle OEF) so

$m\angle DOF=2m\angle OEF=64^\circ$

so the measure of arc DF is also 64 degrees. So we have

$170-2x=64\implies106=2x\implies\boxed{x=53}$

###

2. Arc FE and angle EOF have the same measure, 56 degrees. By the inscribed angle theorem,

$m\angle EOF=2m\angle EDF\implies56^\circ=2m\angle EDF\implies m\angle EDF=28^\circ$

Triangle DEF is isosceles because FD and ED have the same length, so angles EFD and DEF are congruent. Also, the sum of the interior angles of any triangle is 180 degrees. It follows that

$m\angle EFD+m\angle EDF+m\angle DEF=180^\circ\implies\boxed{m\angle EFD=76^\circ}$