So from 1986 you Subtract total percentage 24.65% an $48 And the answer is a
Answer:
a. Since 10 out of 70 had a tag we can write the ratio 10:70. We need to solve for x in 90:x. x = 630 so the answer is 630.
b. An assumption he could have made is that the number of deer that had a tag and the total amount of deer were directly proportional.
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>
Given data :
a₃ = 9/16
aₓ = -3/4 · aₓ₋₁
Where x is the number of terms ('x' is also written as 'n')
To find the 7th term (a₇):
We know that aₓ = -3/4 · aₓ₋₁
So,
a₃ = -3/4 · a₃₋₁
a₃ = -3/4 · a₂
9/16 = -3/4 · a₂
a₂ = 9/16 × -4/3
a₂ = -36/48
a₂ = -3/4
Again,
aₓ = -3/4 · aₓ₋₁
a₄ = -3/4 · a₄₋₁
a₄ = -3/4 · a₃
a₄ = -3/4 · 9/16
a₄ = -27/64
a₄ = -27/64
For a₅,
aₓ = -3/4 · aₓ₋₁
a₅ = -3/4 · a₅₋₁
a₅ = -3/4 · a₄
a₅ = -3/4 × -27/64
a₅ = 81/256
For a₆,
aₓ = -3/4 · aₓ₋₁
a₆ = -3/4 · a₆₋₁
a₆ = -3/4 · a₅
a₆ = -3/4 × 81/256
a₆ = -243/1024
For a₇,
aₓ = -3/4 · aₓ₋₁
a₇ = -3/4 · a₇₋₁
a₇ = -3/4 · a₆
a₇ = -3/4 × -243/1024
a₇ = 729/4096