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>
Let x = the # jujubes Sam had
he gave x/2 to jen
jen eats half of it = 1/2(x/2) = x/4
the remaining x/4 he gave to kyle
kyle kept 8 and gave 10 to Kim
so
x/4 - 8 = 10
x/4 = 18
x =72
thus jen had 36 jujubes
jen eat 18 jujubes
Given : tan 235 = 2 tan 20 + tan 215
To Find : prove that
Solution:
tan 235 = 2 tan 20 + tan 215
Tan x = Tan (180 + x)
tan 235 = tan ( 180 + 55) = tan55
tan 215 = tan (180 + 35) = tan 35
=> tan 55 = 2tan 20 + tan 35
55 = 20 + 35
=> 20 = 55 - 35
taking Tan both sides
=> Tan 20 = Tan ( 55 - 35)
=> Tan 20 = (Tan55 - Tan35) /(1 + Tan55 . Tan35)
Tan35 = Cot55 = 1/tan55 => Tan55 . Tan35 =1
=> Tan 20 = (Tan 55 - Tan 35) /(1 + 1)
=> Tan 20 = (Tan 55 - Tan 35) /2
=> 2 Tan 20 = Tan 55 - Tan 35
=> 2 Tan 20 + Tan 35 = Tan 55
=> tan 55 = 2tan 20 + tan 35
=> tan 235 = 2tan 20 + tan 215
QED
Hence Proved
Recall that
There are three cases to consider:
(1) When
, we have
and
, so
(2) When
and
, we get
and
, so
(3) When
, we have
and
, so
So
A, the first one only, this parabola only has a minimum and no maximum. the other statements are also just false
