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charle [14.2K]
3 years ago
8

If the fallowing pair represents a direct variation, find the missing value. (2,-4) and (-6, y)

Mathematics
2 answers:
erik [133]3 years ago
5 0
-8y it should be right
Maurinko [17]3 years ago
5 0
Okay well the relationship from 2 and -6 is multiplication by -3. SO if we apply that to -4, y will equal 12.
(2, -4)
(-6, 12)
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Please help! I'll mark you brainliest :D
Grace [21]

Answer:

range: -3<x<7

Step-by-step explanation:

The curve of this function starts at x=-3 and ends at x=7, so this function's range is : -3<x<7

5 0
3 years ago
Three hundred new ones cost $2100. what would be the cost of 120 new ones​
SSSSS [86.1K]

Answer:

840

Step-by-step explanation:

120 is 40% of 300, and 840 is 40% of 2100 is 840.

5 0
3 years ago
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Each month, the census bureau mails survey forms to 250,000 households asking questions about the people living in the household
allsm [11]

Answer:

Option D is correct. All US households are among the population of interest.

Step-by-step explanation:

The Census bureau is a national body in the U.S. and they do the survey of 250000 different households monthly. The surveys go out to all corners of the U.S. depending on which part of the nation is doing the survey next.

It shows that the entire U.S. population is of interest to the census bureau.

5 0
3 years ago
Pls help with 19 tysm its also worth branliest ty if u help
Mariana [72]
I’ll help you out, but what are the numbers in between? Like does it go boy 0, .3, .6, .9, to 1? Or is there a different pattern
5 0
3 years ago
Read 2 more answers
he number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.02
Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2  

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

P(k)= \frac{0.2^ke^{-0.2}}{k!}

a)

In this case, we are looking for P(0)

P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and  

P(k)= \frac{2^ke^{-2}}{k!}

and  

P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%

Hence, the probability that at most one car has no flaws is 27.06%

6 0
3 years ago
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