First using the given two points we can find the slope (m) of the line
Thus the slope of given line is 1.
Using slope and a point (0,1) we can write the equation in point slope form as
y - 1= 1(x-0)
y - 1 = x
The above equation is the point-slope form of the equation.
Answer:
If f(x) = 2x2 - 4x + 3 evaluate for f(3) and f(-4) 2. g(x) = V - 2x; find g(-27) and g(-1) ... {(-3,0), (+3, 3),(-3,2)} ... If f(x) = 5x+5, find and simplify f(4).
Answer: Graph D
Step-by-step explanation: Got it right on plato
For problems like this, change the wording into simple algebraic
formulas. We know that the area of a rectangle is its width times its
length, or A=LW. The problem tells you that the area is 48, so LW=48.
Also, if the length = twice the width minus 4, then L=2W-4.
From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W
Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0.
Solving this equation gives us W values of 6 and -4, but since the width
of a rectangle cannot be negative, the width must be 6.
Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.
To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.
Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.
Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.
Answer: A: 0.0031
Step-by-step explanation:
Given : In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a standard deviation of 5.2 minutes.
i.e. 
and 
We assume that the wait times are normally distributed.
samples size : n= 30
Let x denotes the sample mean wait time.
Then, the probability that the mean wait time is greater than 20 minutes will be :
![P(x>20)=1-P(x\leq20)\\\\=1-P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{20-17.4}{\dfrac{5.2}{\sqrt{30}}})\\\\=1-P(z\leq2.74)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.9969\ \ [\text{ By z table}]\\\\=0.0031](https://tex.z-dn.net/?f=P%28x%3E20%29%3D1-P%28x%5Cleq20%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5Cleq%5Cdfrac%7B20-17.4%7D%7B%5Cdfrac%7B5.2%7D%7B%5Csqrt%7B30%7D%7D%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.74%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%5D%5C%5C%5C%5C%3D1-0.9969%5C%20%5C%20%5B%5Ctext%7B%20By%20z%20table%7D%5D%5C%5C%5C%5C%3D0.0031)
Hence, the probability that the mean wait time is greater than 20 minutes.= 0.0031
Thus , the correct answer is A: 0.0031 .
