Answer:
x = -1 and x = 5
Step-by-step explanation:
<em>What are the solutions of the equation (x – 3)² + 2(x – 3) -8 = 0? Use u substitution to solve.</em>
<em />
(x – 3)² + 2(x – 3) -8 = 0 -------------------------------------------------------(1)
To solve this problem, we will follow the steps below;
let u = x-3
we will replace x-3 by u in the given equation:
(x – 3)² + 2(x – 3) -8 = 0
u² + 2u -8 = 0 ----------------------------------------------------------- --------------(2)
We will now solve the above quadratic equation
find two numbers such that its product gives -8 and its sum gives 2
The two numbers are 4 and -2
That is; 4×-2 = -8 and 4+(-2) = 2
we will replace 2u by (4u -2u) in equation (2)
u² + 2u -8 = 0
u² + 4u - 2u -8 = 0
u(u+4) -2(u+4) = 0
(u+4)(u-2) = 0
Either u + 4 = 0
u = -4
or
u-2 = 0
u = 2
Either u = -4 or u = 2
But u = x-3
x = u +3
when u = -4
x = u + 3
x = -4 + 3
x=-1
when u = 2
x = u + 3
x = 2 + 3
x=5
Therefore, x = -1 and x =5
x
Answer:
- <u><em>Option b. just below 30%</em></u>
<u><em></em></u>
Explanation:
Please, see attached the <em>histogram that represents the distribution of acceptance rates (percent accepted) among 25 business schools in 2004. </em>
<em />
The<em> median</em> is the value that separates the lower 50% from the upper 50% of the data.
Since there are 25 business schools, the middle value is the number 13.
The height of each bar is the<em> frequency</em> or number of business school for that acceptace rate:
- The first bar has frequency of 1 school
- The second bar has frequency of 3 schools: cummulative frequency: 1+3=4.
- The third bar has frequency 5 schools: cummulative frequency 4 + 5 = 9.
- The fourth bar has frequency 3 schools: cummulative frequency: 9+3=12.
Then, the 13th value is on the next bar, the fifth bar.
The fifth bar has acceptance rates 25 ≤ rate < 30.
That means that the median acceptance rate is greater than or equal to 25 and less than 30.
Thus, the choice is the option <em>b. just below 30%.</em>
Answer:
Step-by-step explanation:
a)
Test statistic:




here test statistic lie in rejection region,that why null hypothesis fails
so Yes, its significant.
b)
Test statistic:




c)
sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.
Average rate of change = [H(100) - H(80)] / (100 - 80)
H(100) = 0.003(100)^2 + 0.07(100) - 0.027 = 0.003(10000) + 0.07(100) - 0.027 = 30 + 7 - 0.027 = 36.973
H(80) = 0.003(80)^2 + 0.07(80) - 0.027 = 0.003(6400) + 0.07(80) - 0.027 = 19.2 + 5.6 - 0.027 = 24.773
Average rate of change = (36.973 - 24.773)/(100 - 80) = 12.2/20 = 0.61
Answer: B