Write all the factors of 12
1 answer:
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Answer:
D) As the gas mileage increases, the cost of driving a fixed distance increases.
Step-by-step explanation:
Hello!
Given the variables
X₁: Gas mileage of a car
X₂: Cost of driving 500 miles
Normally the variable that changes more is the gas mileage of the car, depending on the type of the car or the age of different cars from the same model and brand, the amount of gas used to cover a fixed distance varies.
From a logical point of view you'd expect that the more gas the car consumes, the more it will cost to drive a fixed distance. So the correct option is D)
Options:
A) As the gas mileage increases, the cost of driving a fixed distance decreases.
B) As the cost of driving a fixed distance increases, the gas mileage decreases.
C) As the cost of driving a fixed distance decreases, the gas mileage increases.
D) As the gas mileage increases, the cost of driving a fixed distance increases.
I hope this helps!
1. The first step here is to arrange the data set's form least to greatest,
Sherelle: 26, 39, 56, 58, 60, 62, 65, 66, 66, 68, 71, 72, 72, 73, 74, 75, 81, 83, 84, 85
Venita: 44, 45, 51, 51, 53, 53, 55, 57, 58, 62, 65, 66, 69, 69, 70, 73, 75, 77, 78, 79
Now we can determine our 5 - number summary based on the numbers respective positions.
First Data Set,
<u>(Five - Number Summary) - Minimum : 26, Quartile 1 : 60, Median : 69.5, Quartile 3 : 75, Maximum : 85</u>
Second Data Set,
<u>(Five - Number Summary) - Minimum : 44, Quartile 1 : 53, Median : 63.5, Quartile 3 : 73, Maximum : 79</u>
2. This part is based on your drawings of the box and whisker plots, so you would have to figure that part out by yourself.
3. First off we know that our data set is composed of the years from 1900, so let's rewrite the set based off of the actual year -
Sherelle: 1926, 1939, 1956, 1958, 1960, 1962, 1965, 1966, 1966, 1968, 1971, 1972, 1972, 1973, 1974, 1975, 1981, 1983, 1984, 1985
Venita: 1944, 1945, 1951, 1951, 1953, 1953, 1955, 1957, 1958, 1962, 1965, 1966, 1969, 1969, 1970, 1973, 1975, 1977, 1978, 1979
( a ) Now in Sherelle's defence, she can say that the lowest coin date in her group is 1926, comparative to Venita's group - the lowest coin date in hers being 1944. Therefore, she is more likely to have the 1916 coin, after all that date is the lowest overall in both their data set.
( b ) In Venita defence, she can say that the mean of her data set is lower than the mean of Sherelle's data set. Take a look at the calculations below,
Sherella's Mean : =
= 1966.8,
Venita's Mean : =
= 1962.5
( c ) I would say Sherella's bag would most likely contain the 1916 coin. The mean is a prominent factor, but their mean(s) only differ by a very small quantity. That too, Sherella's bag contains the lowest coin in both their groups, and though that is not a prominent factor, it could be that she does have the 1916 coin.
