Answer:
(4,2.5)
.................................
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
-15v-40=23-8v
We're trying to isolate the v so...
1) -15v= 63-8v add the 40 on both sides
2) -7v=63 add the 8v on both sides
3) divide -7 on both sides to solve for v
4) v= -9
Based on the one-sample t-test that Mark is using, the two true statements are:
- c.)The value for the degrees of freedom for Mark's sample population is five.
- d.)The t-distribution that Mark uses has thicker tails than a standard normal distribution.
<h3>What are the degrees of freedom?</h3><h3 />
The number of subjects in the data given by Mark is 6 subjects.
The degrees of freedom can be found as:
= n - 1
= 6 - 1
= 5
This is a low degrees of freedom and one characteristic of low degrees of freedom is that their tails are shorter and thicker when compared to standard normal distributions.
Options for this question are:
- a.)The t-distribution that Mark uses has thinner tails than a standard distribution.
- b.)Mark would use the population standard deviation to calculate a t-distribution.
- c.)The value for the degrees of freedom for Mark's sample population is five.
- d.)The t-distribution that Mark uses has thicker tails than a standard normal distribution.
- e.)The value for the degrees of freedom for Mark's sample population is six.
Find out more on the degrees of freedom at brainly.com/question/17305237
#SPJ1
