The answer is A they both get payed 6 dollars per hour
Answer:
Ivy is correct
Step-by-step explanation:
the perimeter of a parallelogram can be written as P= 2(a+b), where a and b are the lengths of the sides of the parallelogram.
since the parallelogram has been dilated by a scale factor of 0.6, we multiple the original side lengths by 0.6.
12 × 0.6 = 7.2
16 × 0.6 = 9.6
Plug into the equation
P=2(7.2+9.6)
P=2(16.8)
P= 33.6
You can also solve by plugging in the original values and multiplying the answer by 0.6.
P=2(12+16)(0.6)
P=2(28)(0.6)
P=(56)(0.6)
P=33.6
It's not a function because the x value repeated but wasn't accompanied by the same y value . there is (-2,2) and (-2,14) but if they had the same y value like (-2,2) and (-2,2) or (-2,14) and (-2,14) the table would represent a function. If you still dont understand I'll be more than happy to elaborate :)
Answer:
4/9
Step-by-step explanation:
The possibilities of transportation: (The first will be for morning, second will be for afternoon)
B, B
B, C
B, T
C, B
C, C
C, T
T, B
T, C
T, T
It is clearly seen that there are 9 transportation options.
(Using cab 1 time we have BC, CB, CT, TC. So four of the transportation methods use cab one time.)
Therefore, the probability that she will use a cab only once is 4/9.
The base case is the claim that

which reduces to

which is true.
Assume that the inequality holds for <em>n</em> = <em>k </em>; that

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

By the induction hypothesis,

Now compare this to the upper bound we seek:

because

in turn because
