Answer:
Table a match with Graph 3
Table b match with Graph 2
Table c match with Graph 4
Step-by-step explanation:
<u>Table a</u>
Time is plotted on x - axis and Temp on y axis
From table a we can see that the y values(temp) decrease with increasing time. The only graph that has a decreasing trend is Graph 3
<u>Table b</u>
Time plotted on x axis, cost on y axis
Looking at the values we see that they are almost linear except for 2 values at x = 2 and 2.5 and at y = 5 and 5.3.
Ignoring values 2.5 and 5.3 we see a linear fit with a slope of 20. This means either graph 1 or graph 2
However, the graph passes through (0,0) and this is not a set of values in the table. That leaves graph 2
<u>Table c</u>
Months plotted on x axis and length of fetus on the y axis
We can see that the y values (the length) increases slowly at lower values of x(x=1 and x=2) and then increase rapidly at the mid values(x = 2 thru 6) and then slows down between months 6 and 9. Only Graph 4 fits this pattern
4 is your answer have a great day
Volume is legnth times widht times height
lenght=2x-1
width=x-2
height=x+1
multiply all together
use mass distributive property
distributive=a(b+c)=ab+ac so extending that
(a+c)(c+c)=(a+b)(c)+(a+b)(d) then keep distributing so
(2x-1)(x-2)(x+1)
do each one seperately
do the first two first and put the other one (x+1) to the side for later
(2x-1)(x-2)=(2x-1)(x)+(2x-1)(-2)=(2x^2-x)+(-4x+2)=2x^2-5x+2
then do the other one
(x+1)(2x^2-5x+2)=(x)(2x^2-5x+2)+(1)(2x^2-5x+2)=(2x^3-5x^2+2x)+(2x^2-5x+2)=2x^3-3x^2-3x+2
the lasst form is 2x^3-3x^2-3x+2
First, we need to work out the total number of students who were being surveyed.
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let

be the number of pets and

is the probability of owning the number of pets