The product of the given two matrices comes out to be ![\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Here we are given the 2 matrices as follows-
![\left[\begin{array}{ccc}7&-2\\-6&2\end{array}\right] \left[\begin{array}{ccc}1&1\\3&3.5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-2%5C%5C-6%262%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C3%263.5%5Cend%7Barray%7D%5Cright%5D)
To find the product of 2 matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
Here since both of the matrices are 2 × 2, their product is possible.
Now, to find the product, we need to multiply each element in the first row by each element of the 1st column of the second matrix and then find their sum. Similarly, we do this for all rows and columns.
Therefore,
![\left[\begin{array}{ccc}(7*1)+(-2*3)&(7*1)+(-2*3.5)\\(-6*1)+(2*3)&(-6*1)+(2*3.5)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%287%2A1%29%2B%28-2%2A3%29%26%287%2A1%29%2B%28-2%2A3.5%29%5C%5C%28-6%2A1%29%2B%282%2A3%29%26%28-6%2A1%29%2B%282%2A3.5%29%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}(7)+(-6)&(7)+(-7)\\(-6)+(6)&(-6)+(7)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%287%29%2B%28-6%29%26%287%29%2B%28-7%29%5C%5C%28-6%29%2B%286%29%26%28-6%29%2B%287%29%5Cend%7Barray%7D%5Cright%5D)
=
Thus, the product of the given two matrices comes out to be
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Answer:
297 cards in a year
Step-by-step explanation:
There are 12 months in a year, and he gets 2 packs of cards every month.
12*2 = 24 packs of cards a year
He gets 3 extra packs of cards for his birthday.
24 + 3 = 27 packs of cards
There are 11 cards in each pack.
27 X 11 = 297 cards in a year
Answer:22%
Step-by-step explanation:
In time period of 1980-1988 the rate of ticket price is $0.2 per year
Between time period 1989-1993 there is constant rate.
Between year 1994-2011 the increase in rate is same $0.2 per year
The graph could be divided up into three different periods of relatively consistent ticket price change: The years 1980 – 1988, 1989 – 1993 and 1994 – 2011.
<h3>What is Statistic?</h3>
The statistic is the study of mathematics which deal with relations between comprehensive data.
The graph is not available, in the question, so the graph could be as attached
For period 1980-1988
rate of change = 4.2-2.8/ 8 = 0.2
In time period of 1980-1988 the rate of ticket price is $0.2 per year
for period 1989-92 there is a straight line so,
Between time period 1989-1993 there at constant rate.
For period, 1994-2011
rate of change = 4.4-8/17 = 0.2
Between year 1994-2011 the increase in rate is same $0.2 per year
Thus, for the 3 Time period we have rate of change in ticket price is $0.2 per year, no change in ticket price, $0.2 per year respectively.
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