9514 1404 393
Answer:
A) Enterprise: E(x) = 50 +0.25x; Hertz: H(x) = 30 +0.45x
B) see attached graph
Step-by-step explanation:
A) We choose to compare Hertz (lowest daily charge) and Enterprise (lowest mileage charge). In each case, the total charge will be the sum of the daily charge and the product of the mileage charge and number of miles (x).
Enterprise: E(x) = 50 +0.25x
Hertz: H(x) = 30 +0.45x
Answer:
Probability that Caroline buys fruit, a CD or both is 0.76.
Step-by-step explanation:
Let event A = Caroline buys fruit, event B = Caroline buys CD, Ac and Bc are complementary events.
Events AB, ABc, AcB and AcBc are jointly exhaustive and disjoint, hence P(AB) + P(ABc) + P(AcB) +P(AcBc) =1.
Events A and B independent, hence Ac and Bc independent too and probability P(AcBc) = P(Ac)*P(Bc) = (1 - P(A))(1-P(B)) = 0.6*0.4 = 0.24.
Required probability P(AB + ABc + AcB ) = P(AB) + P(ABc) + P(AcB) = 1- P(AcBc) = 1 - 0.24 = 0.76.
Data:
15 16 14 15 19 17
n=6 points
sum is 96
mean is 96/6 = 16
Now we look at the absolute deviations, each of which is the absolute value of a score minus the mean, basically the distance of the score to the mean .
Scores 15 16 14 15 19 17
AbsDev 1 0 2 1 3 1
The sum of the absolute deviations is 8 and there are six of them so the
Mean Absolute Deviation = 8/6 = 4/3
Answer: 2. 8/6
<span>A. Lines that go up from left to right should have a positive slope.
B. Lines that go down from left to right should have a negative slope.
</span><span>D. The slope of a steep line should be bigger than the slope of a flat line.
E. The slope of a steep line should be smaller than the slope of a flat line.</span>
These array of numbers shown above are called matrices. These are rectangular arrays of number that are arranged in columns and rows. It is mostly useful in solving a system of linear equations. For example, you have these equations
x+3y=5
2x+y=1
x+y=10
In matrix form that would be
![\left[\begin{array}{ccc}1&3&5\\2&1&1\\1&1&10\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%265%5C%5C2%261%261%5C%5C1%261%2610%5Cend%7Barray%7D%5Cright%5D%20)
where the first column are the coefficients of x, the second column the coefficients of y and the third column is the constants, When you multiple matrices, just multiply the same number on the same column number and the same row number. For this problem, the solution is
![\left[\begin{array}{ccc}3*2&0*8\\2*0.6&-1*3\end{array}\right] = \left[\begin{array}{ccc}6&0\\1.2&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%2A2%260%2A8%5C%5C2%2A0.6%26-1%2A3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%5C%5C1.2%26-3%5Cend%7Barray%7D%5Cright%5D%20)
