A function f(x) = x^2 - 4 has a domain as the set of integers
1 answer:
Answer:
8
Step-by-step explanation:
f(x) = (x+2)(x-2)
Substituting x=1 into f(x):
f(1) = 3*-1
= -3
Since -2 & 2 touch the x-axis, 1 has a negative y value, 2 has a y value of 0, and all values from 2, are positive.
Hence,
3, 4, 5, 6, 7, 8, 9, 10 are all positive. Therefore, the answer is 8
This can be confirmed with a graph, as attached below.
<em>Feel free to mark this as brainliest! :D</em>
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Answer:
(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is 0.1271.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.
(e) The time is 5.65 minutes.
Step-by-step explanation:
We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
Let X = <u><em>the length of the calls, in minutes.</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time = 4.5 minutes
= standard deviation = 0.7 minutes
(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X 4.50 min)
P(X < 5.30 min) = P( <
) = P(Z < 1.14) = 0.8729
P(X 4.50 min) = P(
) = P(Z
0) = 0.50
The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.
(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)
P(X > 5.30 min) = P( >
) = P(Z > 1.14) = 1 - P(Z
1.14)
= 1 - 0.8729 = <u>0.1271</u>
The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X 5.30 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 5.30 min) = P(
) = P(Z
1.14) = 0.8729
The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X 4.00 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 4.00 min) = P(
) = P(Z
-0.71) = 1 - P(Z < 0.71)
= 1 - 0.7612 = 0.2388
The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.
(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;
P(X > x) = 0.05 {where x is the required time}
P( >
) = 0.05
P(Z > ) = 0.05
Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;
x = 4.5 + 1.15 = 5.65 minutes.
SO, the time is 5.65 minutes.
Answer:
- 25000 seats in section A
- 14100 seats in sectin B
- 10900 seats in section C
Step-by-step explanation:
The problem statement tells you half the total number of seats are in section A, so you already know that there are 25000 A seats. The revenue from those seats is
... 25000×$25 = $625,000
so the revenue from B and C seats must total
... $1,070,500 - 625,000 = $445,500
If all 25000 of the B/C seats were C seats, the revenue would be
... 25000×$15 = $375,000
The actual revenue from those seats is $445,500 -375,000 = $70,500 more than that. We know each B seat generates $5 more revenue, so there must be ...
... $70,500/$5 = 14,100 . . . . B seats
Then the balance of the 25000 B and C seats are C seats:
... 25,000 - 14,100 = 10,900 . . . . C seats
_____
<em>Alternate Solution Method</em>
The new Brainly answer format requires the answer be supplied before the working. In order to find the answer quickly so that I can fill in that section, I used a matrix method for solving the problem. The problem equations can be written ...
- a + b + c = 50000
- a - b - c = 0
- 25a + 20b + 15c = 1070500
so the augmented matrix is ...
A graphing calculator can be used to find the solution to this, generally using a function that produces the reduced row-echelon form. The attachment shows the solution using a TI-84 calculator.
___
<em>Comment on the Working</em>
Since the number of A seats is equal to the total of B and C seats, the number of A seats must be half the total number of stadium seats. Having figured that out, the problem is reduced to one of finding the mix of B and C seats that will produce the remaining revenue.
As with many mixture problems, it is convenient to look at differences. Start with the assumption that all of the desired revenue comes from the least contributor. Here, that is C seats. Then figure the difference that using a B seat makes ($20 -15 = $5) and the difference of the actual revenue and the amount that you got by assuming all C seats: 445,500 -375,000 = 70,500. Since replacing a C seat by a B seat adds $5 to the revenue, it is easy to figure the number of such replacements required in order to raise the revenue by $70,500.
If you write the equation for B seats, you find the solution to the equation mirrors this verbal description:
... 20b + 15(25000-b) = 445,500
... 5b = 445,500 - 375000 . . . . simplify, subtract 375000
... b = 70500/5 = 14100
