Hello!
To write this in slope intercept form we have to solve for y.
Add 3x to both sides.
9y=-27+3x
Divide both sides by 9
y=-3+1/3
Rearrange into slope intercept form.
y=1/3x-3
I hope this helps!
When x=2 y = 110 which means when when x=1 y = 55
the graph would be 11*55 or 605
The table would be 280/4= 70 70*11=770
Which means the difference is 770-605=165
165 is ur answer
Divide The 40 By 5 Than Multiply The Answer By 2.
So The Answer Would Be <u>16</u>
Answer: 0.2789
Step-by-step explanation:
Given: Mean : ![\mu=10.00\ mm](https://tex.z-dn.net/?f=%5Cmu%3D10.00%5C%20mm%20)
Standard deviation : ![\sigma =0.03\ mm](https://tex.z-dn.net/?f=%5Csigma%20%3D0.03%5C%20mm)
The formula to calculate z-score is given by :_
![z=\dfrac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
For x= 9.96 mm, we have
![z=\dfrac{10-9.96}{0.03}\approx1.33](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B10-9.96%7D%7B0.03%7D%5Capprox1.33)
For x= 10.01 mm, we have
![z=\dfrac{10.01-10}{0.03}\approx0.33](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B10.01-10%7D%7B0.03%7D%5Capprox0.33)
The P-value = ![P(0.33](https://tex.z-dn.net/?f=P%280.33%3Cz%3C1.33%29%3DP%28z%3C1.33%29-P%28z%3C0.33%29)
![= 0.9082408- 0.6293=0.2789408\approx0.2789](https://tex.z-dn.net/?f=%3D%200.9082408-%200.6293%3D0.2789408%5Capprox0.2789)
Hence, the probability that a randomly selected gear has a diameter between 9.96 mm and 10.01 mm =0.2789
Elouise and Marcus are 20 km apart. There is a lake due west of Elouise that is also 10√3 km due north of Marcus, so 22.36 km distance the bearing of Marcus from Elouise.
<h3>What is Pythagoras theorem?</h3>
According to the Pythagoras theorem, the square of the hypotenuse of a right-angled triangle equals the sum of the squares of the other two sides. The formula for this theorem is c² = a² + b², where c is the hypotenuse and a and b are the triangle's two legs. Pythagoras theory triangles are another name for these triangles.
At first assume a triangle, In the triangle, assume Elouise at the point A and Marcus at the point B, and the lake at the C point.
Given that,
Elouise and Marcus are 20 km apart (AB)
There is a lake due west of Elouise that is also 10√3 km due north of Marcus (BC)
Now, according to a Pythagoras theorem,
(AC)² + (AB)² = (BC)²
or, (AC)² = (BC)²- (AB)²
or, (AC)² = (10√3)² - (20)²
or, AC = ![\sqrt{500}](https://tex.z-dn.net/?f=%5Csqrt%7B500%7D)
or, AC = 22.36 km
So, distance bearing of Marcus from Elouise is 22.36 km.
To know more about distance refer to:
brainly.com/question/13963928
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