The conclusion is P(pass l morning) = 0.56
P(pass l afternoon) = 0.72
Conclusion: a student taking the test in the afternoon has a greater probability of passing than a student taking the test in the morning.
<h3>What are the probabilities?</h3>
Probability is the odds that a random event would happen. The odds that the event would happen lies between 0 and 1. The closer the probability is to one, the more likely it is for the event to happen.
P(pass l morning) = number of students who took the test in the morning and passed / total number of students that took the exam in the morning
28 / 50 = 0.56
P(pass l afternoon) = number of students who took the test in the afternoon and passed / total number of students that took the exam in the afternoon = 36 / 50 = 0.72
To learn more about probability, please check: brainly.com/question/13234031
#SPJ1
Scale factor = new length/old length
scale factor = 10/2.5 = 4
The scale factor is 4.
Below are suppose the be the questions:
a. factor the equation
<span>b. graph the parabola </span>
<span>c. identify the vertex minimum or maximum of the parabola </span>
<span>d. solve the equation using the quadratic formula
</span>
below are the answers:
Vertex form is most helpful for all of these tasks.
<span>Let </span>
<span>.. f(x) = a(x -h) +k ... the function written in vertex form. </span>
<span>a) Factor: </span>
<span>.. (x -h +√(-k/a)) * (x -h -√(-k/a)) </span>
<span>b) Graph: </span>
<span>.. It is a graph of y=x^2 with the vertex translated to (h, k) and vertically stretched by a factor of "a". </span>
<span>c) Vertex and Extreme: </span>
<span>.. The vertex is (h, k). It is a maximum if "a" is negative; a minimum otherwise. </span>
<span>d) Solutions: </span>
<span>.. The quadratic formula is based on the notion of completing the square. In vertex form, the square is already completed, so the roots are </span>
<span>.. x = h ± √(-k/a)</span>
notice... the dog's pen perimeter, does not include the side that's bordering the garden's, since that side will use the heavy duty fence, instead of the light one
so, the sum of both of those costs, will be the C(x)
so, just take the derivative of it, and set it to 0 to find the extremas, and do a first-derivative test for any minimum
