The change in the water vapors is modeled by the polynomial function c(x). In order to find the x-intercepts of a polynomial we set it equal to zero and solve for the values of x. The resulting values of x are the x-intercepts of the polynomial.
Once we have the x-intercepts we know the points where the graph crosses the x-axes. From the degree of the polynomial we can visualize the end behavior of the graph and using the values of maxima and minima a rough sketch can be plotted.
Let the polynomial function be c(x) = x
² -7x + 10
To find the x-intercepts we set the polynomial equal to zero and solve for x as shown below:
x
² -7x + 10 = 0
Factorizing the middle term, we get:
x
² - 2x - 5x + 10 = 0
x(x - 2) - 5(x - 2) =0
(x - 2)(x - 5)=0
x - 2 = 0 ⇒ x=2
x - 5 = 0 ⇒ x=5
Thus the x-intercept of our polynomial are 2 and 5. Since the polynomial is of degree 2 and has positive leading coefficient, its shape will be a parabola opening in upward direction. The graph will have a minimum point but no maximum if the domain is not specified. The minimum points occurs at the midpoint of the two x-intercepts. So the minimum point will occur at x=3.5. Using x=3.5 the value of the minimum point can be found. Using all this data a rough sketch of the polynomial can be constructed. The figure attached below shows the graph of our polynomial.
Okay, so we know a number of students and the amount of girls as a bracket. So what we have to do now is multiply them with each other.
28 *4/7 = 16
So there are 16 girls in Mr. Chang´s class.
You can also determine the boys by - the students with the girls:
28-16=12 boys
Have a nice day :D
Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Answer:A
Step-by-step explanation:
we will check each options
option-A:
For solving any equations , we always isolate variables on anyone side
For exp: x+7=1
so, this is TRUE
option-B:
For solving system of equations
For exp:
x-y=1
x+y=3
If we use addition , we could easily solve for x and y
so, this is TRUE
option-C:
We always solve problems using conventional method
we do not guess
so, this is FALSE
option-D:
We often reverse order of operation
For exp:
(x-2)^2-3=0
so, this is TRUE
option-E:
For linear equations , we always get one solution , infinite solutions or no solutions
so, this is FALSE
