Answer:
![\sf y= \dfrac{1}{2}x-5](https://tex.z-dn.net/?f=%5Csf%20y%3D%20%5Cdfrac%7B1%7D%7B2%7Dx-5)
Explanation:
- coordinates taken: (0, -5), (6, -2)
slope:
![\rightarrow \sf \dfrac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
![\rightarrow \sf \dfrac{-2--5}{6-0}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20%5Cdfrac%7B-2--5%7D%7B6-0%7D)
![\rightarrow \sf \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20%5Cdfrac%7B1%7D%7B2%7D)
<u>equation in slope intercept form:</u>
- y = m(x) + b [ where "m is slope", "b is y-intercept" ]
![\sf y= \dfrac{1}{2}x-5](https://tex.z-dn.net/?f=%5Csf%20y%3D%20%5Cdfrac%7B1%7D%7B2%7Dx-5)
First, you need to figure out how much the boots weigh together. You can do this by subtracting 144.9 from 150.7.
150.7-144.9=5.8
Assuming each boot weights the same, you can divide 5.8 by 2 to find the weight of each boot.
5.8/2=2.9
Each boot weighs 2.9 pounds.
I hope this helps :)
Answer with Step-by-step explanation:
Since we have given that
Average per week in sales = $8000
Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increase the average sales per salesperson
So, the appropriate null and alternate hypothesis would be
![H_0:\mu=8000\\\\H_a:\mu>8000](https://tex.z-dn.net/?f=H_0%3A%5Cmu%3D8000%5C%5C%5C%5CH_a%3A%5Cmu%3E8000)
b. What is the Type I error in this situation? What are the consequences of making this error?
Type 1 error are those errors in which null hypothesis are supposed to be rejected, but it does not get rejected.
It means sales per week is greater than $8000 but in actual it is not.
c. What is the Type II error in this situation? What are the consequences of making this error?
Type 2 are error are those errors in which null hypothesis are supposed to be accepted but it get rejected.
It means average sales per week is actually $8000 but it is calculated that average sales is less than $8000.
The numbers are 105 and 50
<em><u>Solution:</u></em>
Let "x" be the first number
Let "y' be the second number
Twice a number plus twice a second number is 310
Therefore,
twice of x + twice of y = 310
2x + 2y = 310 ---------- eqn 1
The difference between the numbers is 55
x - y = 55 -------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 2,
x = 55 + y ------ eqn 3
<em><u>Substitute eqn 3 in eqn 1</u></em>
2(55 + y) + 2y = 310
110 + 2y + 2y = 310
4y = 310 - 110
4y = 200
<h3>y = 50</h3>
<em><u>Substitute y = 50 in eqn 3</u></em>
x = 55 + 50
<h3>x = 105</h3>
Thus the numbers are 105 and 50
y=3x-2 y=-2x+8
3x-2=-2x+8 (plus 2 each side)
3x=-2x+6 (plus 2x each side)
5x=6 (divide by 5 each side)
x=1.2
Plug x back into an equation
y=3(1.2)-2
y=3.6-2
y=1.6
Answer: (1.2, 1.6)