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BlackZzzverrR [31]
3 years ago
14

Describe an example in which two variables are strongly correlated, but changes in one do not cause changes in the other.

Mathematics
1 answer:
miv72 [106K]3 years ago
6 0

Just because 2 things are related doesn't mean they will both be changed when one thing happens. For example steak and candy, just because they are both food doesn't mean they taste the same or interact with each other.

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It would be x + 5.75 because the child ticket is 5.75 less than the adult ticket so to solve how to get the adult ticket you do the opposite
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Merchandise is ordered on November 10; the merchandise is shipped by the seller and the invoice is prepared, dated, and mailed b
natali 33 [55]

Answer:

November 13

Step-by-step explanation:

Following dates are given

On November 10 = Merchandise ordered

Date of an invoice prepared, dated and mailed = November 13

Date when the merchandised received by the buyer = November 18

So, the credit period begins when the invoice is prepared, dated and the mailed by the seller to the buyer as it is the evidence of that the merchandise is ordered            

4 0
3 years ago
The freezing point of water is 32°F. When water is at or below this temperature, it is in solid form. Which statements are true
mamaluj [8]

Answer:

The number 32 is part of the solution set

The circle on the graph is closed

The arrow on the graph points to the left

Step-by-step explanation:

The number 32 is part of the solution because it said that water freezes at 32 degrees when it is at or below 32

The circle on the graph is closed because 32 is part of the answer , when the solution is showed then the circle is closed.

Last but not least the arrow on the graph points left because water freezes when it is at 32 or below 32 degrees, so when it points left the numbers become lower.

Hope this helped! :)

8 0
3 years ago
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The pair of figures are simular.find the value of each variable
kkurt [141]
Y=12 because the triangles have proportional side lengths
5 0
3 years ago
The miles-per-gallon rating of passenger cars is a normally distributed random variable with a mean of 33.8 mpg and a standard d
EleoNora [17]

Answer:

The probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

Step-by-step explanation:

Let the random variable <em>X</em> represent the miles-per-gallon rating of passenger cars.

It is provided that X\sim N(\mu=33.8,\ \sigma^{2}=3.5^{2}).

Compute the probability that a randomly selected passenger car gets more than 37.3 mpg as follows:

P(X>37.3)=P(\frac{X-\mu}{\sigma}>\frac{37.3-33.8}{3.5})

                   =P(Z>1)\\\\=1-P(Z

Thus, the probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

7 0
3 years ago
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