Answer:
12
Step-by-step explanation:
Three cars each need 4 wheels, so that is...
4 + 4 + 4 wheels, which is 12 wheels
Answer:
-2
Step-by-step explanation:
Therefore, the average rate of change over the interval [-4,-1] is -2.
<h3>
<u>Answer:</u></h3>
<h3>
<u>Step-by-step explanation:</u></h3>
Here , two circles are given which are concentric. The radius of larger circle is 10cm and that of smaller circle is 4cm . And we need to find thelarea of shaded region.
From the figure it's clear that the area of shaded region will be the difference of areas of two circles.
Let the,
- Radius of smaller circle be r .
- Radius of smaller circle be r .
- Area of shaded region be
![\bf \implies Area_{Shaded}= Area_{bigger}-Area_{smaller} \\\\\bf\implies Area_{Shaded} = \pi R^2 - \pi r^2 \\\\\bf\implies Area_{shaded} = \pi ( R^2-r^2) \\\\\bf\implies Area_{shaded} = \pi [ (10cm)^2 - (4cm)^2] \\\\\bf\implies Area_{shaded} = \pi [ 100cm^2-16cm^2] \\\\\bf\implies Area_{shaded} = \pi \times 84cm^2 \\\\\bf\implies Area_{shaded} = \dfrac{22}{7}\times 84cm^2 \\\\\bf\implies \boxed{\red{\bf Area_{shaded} = 264 cm^2 }}](https://tex.z-dn.net/?f=%5Cbf%20%5Cimplies%20Area_%7BShaded%7D%3D%20Area_%7Bbigger%7D-Area_%7Bsmaller%7D%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7BShaded%7D%20%3D%20%5Cpi%20R%5E2%20-%20%5Cpi%20r%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%3D%20%5Cpi%20%28%20R%5E2-r%5E2%29%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%3D%20%5Cpi%20%5B%20%2810cm%29%5E2%20-%20%284cm%29%5E2%5D%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cpi%20%5B%20100cm%5E2-16cm%5E2%5D%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cpi%20%5Ctimes%2084cm%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20Area_%7Bshaded%7D%20%20%3D%20%5Cdfrac%7B22%7D%7B7%7D%5Ctimes%2084cm%5E2%20%20%5C%5C%5C%5C%5Cbf%5Cimplies%20%5Cboxed%7B%5Cred%7B%5Cbf%20Area_%7Bshaded%7D%20%3D%20264%20cm%5E2%20%7D%7D)
<h3>
<u>Hence </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> the</u><u> </u><u>shaded </u><u>region</u><u> is</u><u> </u><u>2</u><u>6</u><u>4</u><u> </u><u>cm²</u><u>.</u></h3>
Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.
Total no. of times owner-occupied units had a water supply stoppage lasting 6 or more hours
547000 + 5012000 + 6110000 + 2544000 + 557000 = 14770000
x = no. of times owner-occupied units had a water supply stoppage.
<h3>What is the probability at x =0?</h3>
P(x=0) = 547000/14770000
P(x=0) = 0.0370
Similarly, we have at x=1
P(x=1) = 5012000/14770000
P(x=1) = 0.3393
P(x=2) = 6110000/14770000
P(x=2) = 0.4137
P(x=3) = 2544000/14770000
P(x=3) = 0.1722
P(x=4) = 557000/14770000
P(x=4) = 0.0378
x f(x)
0 0.0370
1 0.3393
2 0.4137
3 0.1722
4 0.0378
Total 1
Observations can you make from a comparison of the number of water supply stoppages reported by owner-occupied units versus renter-occupied units 1.
To learn more about the probability distribution visit:
brainly.com/question/24756209
#SPJ1
X is 5. Hope this helps:)
