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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Answer:
B (-3)
Step-by-step explanation:
By factoring, we can get 4(x-y)
Answer:
a. no, there are no sufficient data to use
b. assumption b is right
Step-by-step explanation:
a. for the standard deviation to be calculated at 99% confidence, the data needed should be the mean (given), the variate value to be calculated for(not given), the standard deviation(not given), the z value of the normal distribution ( not given),
since there are too many unknown, the data given are not sufficient to calculate.
b. validity of this interval requires that coating layer thickness be at least approximately normally distributed.
Let
x-------> one number
y--------> another number
we know that
x=6y--------> equation 1
(1/x)+(1/y)=14/3-----> (x+y)/xy=14/3-----> 3*(x+y)=14xy-----> equation 2
<span>I substitute 1 in 2
</span>3*(6y+y)=14(6y)y------> 21y=84y²-----> y=0.25
x=6y-----> x=6*0.25-----> x=1.5
the answer is
the numbers are 1.5 and 0.25
