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.
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The x would equal 0 and the y would equal 3
Answer:
$7.43
Step-by-step explanation:
f (fries) = 1.01
d (drinks) = 1.07
1.05f + 1.07d = ?
To solve this problem, you need to replace the variable with the amount of each item.
1.05(3) + 1.07(4) = ?
Now we can solve by multiplying and adding:
3.15 + 4.28 = ?
<u>7.43</u>
Answer:
y=6x+21
Step-by-step explanation:
simply 3x2+6x+15
Answer:
i think the values for a and b are: a=2 and b=1/4
